## Concave and Convex Mirrors Simulation

#### QUIZ

Q1) For a spherical concave mirror, virtual images are formed when the object is located

A. between F and C
B. beyond C
C. at C
D. inside F

Q2) Which of the following is not true when an image is formed by an object located between C and F of a concave mirror?
A. Negative magnification
B. Negative image distance
C. Inverted image
D. Enlarged image

#### Concave and Convex Mirrors Simulation

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If you look at the surface of a shiny spoon, you will notice that your reflection is different from what you see in a plane mirror. The spoon acts as a curved mirror, with one side curved inward and the other curved outward. The properties of curved mirrors and the images that they form depend on the shape of the mirror and the object’s position.

Concave Mirrors
The inside surface of a shiny spoon, the side that holds food, acts as a concave mirror. A concave mirror has a reflective surface, the edges of which curve toward the observer. Properties of a concave mirror depend on how much it is curved. Figure 1 shows how a spherical concave mirror works. In a spherical concave mirror, the mirror is shaped as if it were a section of a hollow sphere with an inner reflective surface. The mirror has the same geometric center, C, and radius of curvature, r, as a sphere of radius, r. The line that includes line segment CM is the principal axis, which is the straight line perpendicular to the surface of the mirror that divides the mirror in half. Point M is the center of the mirror where the principal axis intersects the mirror.

Figure 1. The focal point r of a spherical concave mirror is located halfway between the center of curvature and the mirror surface. Rays entering parallel to the principal axis are reflected to converge at the focal point, F.

When you point the principal axis of a concave mirror toward the Sun, all the rays are reflected through a single point. You can locate this point by moving a sheet of paper toward and away from the mirror until the smallest and sharpest spot of sunlight is focused on the paper. This spot is called the focal point of the mirror, the point where incident light rays that are parallel to the principal axis converge after reflecting from the mirror.
The Sun is a source of parallel light rays because it is very far away. All of the light that comes directly from the Sun must follow almost parallel paths to Earth, just as all of the arrows shot by an archer must follow almost parallel paths to hit within the circle of a bull’s-eye.
When a ray strikes a mirror, it is reflected according to the law of reflection. Figure 1 shows that a ray parallel to the principal axis is reflected and crosses the principal axis at point F, the focal point. F is at the halfway point between M and C. The focal length, f, is the position of the focal point with respect to the mirror along the principal axis and can
be expressed as f r/2. The focal length is positive for a concave mirror.

Figure 2. The real image, as seen by the unaided eye (a). The unaided eye cannot see the real image if it is not in a location to catch the rays that form the image (b). The real image as seen on a white opaque screen (c).

Graphical Method of Finding the Image
You have already drawn rays to follow the path of light that reflects off plane mirrors. This method is even more useful when applied to curved mirrors.
Not only can the location of the image vary, but so can the orientation and size of the image. You can use a ray diagram to determine properties of an image formed by a curved mirror. Figure 2 shows the formation of a real image, an image that is formed by the converging of light rays. The image is inverted and larger than the object. The rays actually converge at the point where the image is located. The point of intersection, I, of the two reflected rays determines the position of the image. You can see the image floating in space if you place your eye so that the rays that form the image fall on your eye, as in Figure 2a. As Figure 2b shows, however, your eye must be oriented so as to see the rays coming from the image location. You cannot look at the image from behind. If you were to place a movie screen at this point, the image would appear on the screen, as shown in Figure 2c. You cannot do this with virtual images.
To more easily understand how ray tracing works with curved mirrors, you can use simple, one-dimensional objects, such as the arrow shown in Figure 3a. A spherical concave mirror produces an inverted real image if the object position, do, is greater than twice the focal length, f. The object is then beyond the center of curvature, C. If the object is placed between the center of curvature and the focal point, F, as shown in Figure 3b, the image is again real and inverted. However, the size of the image is now greater than the size of the object.

Figure 3. When the object is farther from the mirror than C, the image is a real image that is inverted and smaller compared to the object (a). When the object is located between C and F, the real image is inverted, larger than the object, and located beyond C (b).

Mathematical Method of Locating the Image
The spherical mirror model can be used to develop a simple equation for spherical mirrors. You must use the paraxial ray approximation, which states that only rays that are close to and almost parallel with the principal axis are used to form an image. Using this, in combination with the law of reflection, leads to the mirror equation, relating the focal length, f, object position, do, and image position, di, of a spherical mirror.

 Mirror Equation $\ \frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_0}$ The reciprocal of the focal length of a spherical mirror is equal to the sum of the reciprocals of the image position and the object position.

When using this equation to solve problems, it is important to remember that it is only approximately correct. It does not predict spherical aberration, because it uses the paraxial ray approximation. In reality, light coming from an object toward a mirror is diverging, so not all of the light is close to or parallel to the axis. When the mirror diameter is small relative to the radius of curvature to minimize spherical aberration, this
equation predicts image properties more precisely.

Magnification Another property of a spherical mirror is magnification, m, which is how much larger or smaller the image is relative to the object.
In practice, this is a simple ratio of the image height to the object height. Using similar-triangle geometry, this ratio can be written in terms of image position and object position.

 Magnification m $\ m=\frac{h_i}{h_0}=\frac{-d_i}{d_0}$ The magnification of an object by a spherical mirror, defined as the image height divided by the object height, is equal to the negative of the image position, divided by the object position.

Image position is positive for a real image when using the preceding equations. Thus, the magnification is negative, which means that the image is inverted compared to the object. If the object is beyond point C, the absolute value of the magnification for the real image is less than 1.
This means that the image is smaller than the object. If the object is placed between point C and point F, the absolute value of the magnification for the real image is greater than 1. Thus, the image is larger than the object.

Figure 4. When an object is located between the focal point and a spherical concave mirror, a virtual image that is upright and larger compared to the object is formed behind the mirror (a), as shown with the stack of blocks (b). What else do you see in this picture?

Virtual Images with Concave Mirrors
You have seen that as an object approaches the focal point, F, of a concave mirror, the image moves farther away from the mirror. If the object is at the focal point, all reflected rays are parallel. They never meet, therefore, and the image is said to be at infinity, so the object could never be seen. What happens if the object is moved even closer to the mirror? What do you see when you move your face close to a concave mirror?
The image of your face is right-side up and behind the mirror. A concave mirror produces a virtual image if the object is located between the mirror and the focal point, as shown in the ray diagram in Figure 4a. Again, two rays are drawn to locate the image of a point on an object. As before, ray 1 is drawn parallel to the principal axis and reflected through the focal point. Ray 2 is drawn as a line from the point on the object to the mirror, along a line defined by the focal point and the point on the object.
At the mirror, ray 2 is reflected parallel to the principal axis. Note that ray 1 and ray 2 diverge as they leave the mirror, so there cannot be a real image. However, sight lines extended behind the mirror converge, showing that the virtual image forms behind the mirror.
When you use the mirror equation to solve problems involving concave mirrors for which an object is between the mirror and the focal point, you will find that the image position is negative. The magnification equation gives a positive magnification greater than 1, which means that the image is upright and larger compared to the object, like the image in Figure 4b.

Figure 5. A convex mirror always produces virtual images that are upright and smaller compared to the object.

Convex Mirrors
In the first part of this chapter, you learned that the inner surface of a shiny spoon acts as a concave mirror. If you turn the spoon around, the outer surface acts as a convex mirror, a reflective surface with edges that curve away from the observer. What do you see when you look at the back of a spoon? You see an upright, but smaller image of yourself.
Properties of a spherical convex mirror are shown in Figure 5. Rays reflected from a convex mirror always diverge. Thus, convex mirrors form virtual images. Points F and C are behind the mirror. In the mirror equation, f and di are negative numbers because they are both behind the mirror.
The ray diagram in Figure 5 represents how an image is formed by a spherical convex mirror. The figure uses two rays, but remember that there are an infinite number of rays. Ray 1 approaches the mirror parallel to the principal axis. The reflected ray is drawn along a sight line from F through the point where ray 1 strikes the mirror. Ray 2 approaches the mirror on a path that, if extended behind the mirror, would pass through F.
The reflected part of ray 2 and its sight line are parallel to the principal axis. The two reflected rays diverge, and the sight lines intersect behind the mirror at the location of the image. An image produced by a single convex mirror is a virtual image that is upright and smaller compared to the object.
The magnification equation is useful for determining the apparent dimensions of an object as seen in a spherical convex mirror. If you know the diameter of an object, you can multiply by the magnification fraction to see how the diameter changes. You will find that the diameter is smaller,
as are all other dimensions. This is why the objects appear to be farther away than they actually are for convex mirrors.

Figure 6. Convex mirrors produce images that are smaller than the objects. This increases the field of view for observers.

Field of view
It may seem that convex mirrors would have little use because the images that they form are smaller than the objects. However, this property of convex mirrors does have practical uses. By forming smaller images, convex mirrors enlarge the area, or field of view, that an observer sees, as shown in Figure 6. Also, the center of this field of view is visible from any angle of an observer off the principal axis of the mirror;
thus, the field of view is visible from a wide perspective. For this reason, convex mirrors often are used in cars as passenger-side rearview mirrors.

Mirror Comparison
How do the various types of mirrors compare? Table 1 compares the properties of single-mirror systems with objects that are located on the principal axis of the mirror. Virtual images are always behind the mirror, which means that the image position is negative. When the absolute value of a magnification is between zero and one, the image is smaller than the object. A negative magnification means the image is inverted relative to the object. Notice that the single plane mirror and convex mirror produce only virtual images, whereas the concave mirror can produce real images or virtual images. Plane mirrors give simple reflections, and convex mirrors expand the field of view. A concave mirror acts as a magnifier when an object is within the focal length of the mirror.

## Principle of DC Motor Simulation

#### QUIZ

Q1) The conventional theory for current flow is being used to determine the direction of magnetic lines of force. Technician A says that the left-hand rule should be used. Technician B says that the right-hand rule should be used. Which technician is correct?

A. Technician A only
B. Technician B only
C. Both Technicians A and B
D. Neither Technician A nor B

Q2) A current in a magnetic field experiences a force. Which of the following statements is NOT true?

A. The force is at right angles to both the wire and the magnetic field.
B. The greatest force is experienced when the wire is at right angles to the magnetic field.
C. If the wire is parallel to the magnetic field, then no force will be experienced.
D. The Right Hand Thumb Rule can be used to predict which way the force will act.

Fleming's Left Hand Rule is used to predict the direction of a force, a field, or a current

#### Principle of DC Motor Simulation

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#### The Motor Principle

What would life be like without the electric motor? Many devices depend on the electric motor: computers, fans, elevators, car windows, and amusement park rides, to name a few. How did the electric motor come to be?

Moving conductors with Electricity
Oersted’s discovery inspired much interest in electricity and magnetism among other scientists. When English physicist Michael Faraday saw that an electric current in a wire caused a compass needle to move, he was curious to see if the reverse would be true.
Could a magnetic field cause a current-carrying conductor to move? Not only did he succeed in showing that it could, but he was able to make the electrical conductor rotate. In 1821, Faraday supported a bar magnet in a pool of mercury, which is a good conductor of electricity. He then suspended a copper wire alongside the bar magnet, allowing the copper to make contact with the liquid mercury. The wire and the liquid mercury were connected to a power source to complete the circuit. When the circuit was connected, the wire rotated around the magnet. This was the first electric motor (Figure 1).

The copper wire in Faraday’s motor design moved because the magnetic field in the copper wire interacted with the magnetic field of the permanent bar magnet. Let us examine the interaction between the two fields. In Figure 2(a) there are two separate magnetic fields. One magnetic field is from a current-carrying conductor with the conventional current directed into the page. The other magnetic field is from the external magnets. Where the two interacting magnetic field lines are pointed in the same direction there is a repulsion force. Where the two interacting field lines are pointed in opposite directions there is an attraction force. The final result is that the conductor is forced downward, as shown in Figure 2(b).

Figure 2 (a) The magnetic field lines around a current-carrying conductor and two permanent magnets (b) The magnetic fields interact to force the conductor in a downward direction.

The movement of a current-carrying conductor in an external magnetic field is described by the motor principle. The motor principle states that a current-carrying conductor that cuts across external magnetic field lines experiences a force perpendicular to both the magnetic field and the direction of the electric current. The magnitude of this force depends on the magnitude of both the external field and the current, and the angle between the conductor and the magnetic field it cuts across.

Right-Hand rule for the Motor Principle
A third right-hand rule can be used as a tool to determine the direction of force acting on a current-carrying conductor. This time your hand is held flat with your thumb at a right angle to your fingers. The right-hand rule for the motor principle states that if the fingers of your open right hand point in the direction of the external magnetic field and your thumb points in the direction of the conventional current, then your palm faces in the direction of the force on the conductor (Figure 4).

Figure 4 The right-hand rule for the motor principle

The Analog Meter
One of the first practical uses of the motor principle was the development of meters for measuring electrical quantities. The motor principle was used to develop the galvanometer—a sensitive meter for measuring current. The first meters were analog.

Figure 5 (a) An analog meter and (b) a cross-section of an analog meter

Analog means that the reading is shown using a moving needle or pointer on a scale; there is no digital display. In the analog meter shown in Figure 5(a), you can see the looped conductor where the current enters. Note that the current is directed to the positive terminal, through the loop and then out of the negative terminal. The needle is perpendicular to the coil and fixed to it. The needle and the coil are free to rotate on an axle. The spring provides just the right amount of tension and does not let the needle continue forward. The scale is there to provide a spot to take readings from. Looking at the cross-sectional view in Figure 5(b), you can see the current directed into the page on the right side and out of the page on the left side. Using the right-hand rule for the motor principle, you can see that the loop is forced up on the left side and down on the right side. This causes the needle to rotate toward the right side of the scale. The main advantage of analog meters over digital meters is that it is much easier to see the rate at which changes in readings occur.

## Principle of Superposition Simulation

#### Quiz

When two waves overlap, the displacement of the medium is the sum of the displacements of the two individual waves. This is the principle of __________.

A. constructive interference
B. destructive interference
C. standing waves
D. superposition

The displacement due to two waves that pass through the same point in space is the algebraic sum of displacements of the two waves.

#### Principle of Superposition Simulation

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Interference of Waves
On the surface of a lake on a windy day, you will see many complicated wave motions (Figure 1). You will not see a simple wave moving in a particular direction. The water surface appears this way because of the action of many thousands of waves from various directions and with various amplitudes and wavelengths. When waves meet, a new wave is generated in a process called interference. In this section, you will learn what happens when two waves meet and interfere with each other.

Figure 1 The surface waves on this lake are the result of the interference of thousands of waves of different wavelengths and amplitudes. Most of these waves are caused by the wind, but they are also caused by passing boats and ships.

Wave Interference at the Particle Level
Waves are the result of particle vibrations, and that the particles in a medium are connected by forces that behave like small springs. Wave interference is influenced by the behaviour of the particles.
Wave motion is efficient: in most media, little energy is lost as waves move. When waves come together, this efficiency continues. When one wave passes in the vicinity of a particle, the particle moves up and down in an oval path, which allows the wave to move in a specific direction, as shown in Figure 2(a). When a second wave is also present, the vibration of the particle is modified. Th e oval motion of the particles stimulates the next particle in the direction of the wave’s motion to begin vibrating.
When two (or more) waves come together, as shown in Figure 2(b), the particle moves up and down rather than in an oval path because the speeds of the combined waves cancel each other out. Th e motion of the particle allows the waves to pass through each other. Th e waves are not modified, so the amount of energy stays the same. Th us, when two or more waves interact, the particle vibration is such that the direction and energy of each wave are preserved. Aft er the waves have passed through each other, none of their characteristics—wavelength, frequency, and amplitude—change.

Figure 2 (a) The basic motion of a vibrating particle in a travelling wave. (b) When two waves meet, the particle motion is more up and down. The wave characteristics are unchanged after the waves pass through each other.

Constructive and Destructive Interference
When two waves meet, the forces on their particles are added together. If the two waves are in phase (the phase shift between them is zero), then the resulting amplitude is the sum of the two original amplitudes. This is called the principle of superposition: the resulting amplitude of two interfering waves is the sum of the individual amplitudes.
Constructive interference occurs when two or more waves combine to form a wave with an amplitude greater than the amplitudes of the individual waves (Figure 3). Destructive interference occurs when two or more waves that are out of phase combine to form a wave with an amplitude less than at least one of the initial waves (Figure 4).

Figure 3 Constructive interference. Two wave pulses approach each other on a rope. Notice how the amplitudes of the two waves add together. Notice, also, how the waves are unchanged after they pass through each other. The amplitude during interference in (c) is the sum of the amplitudes of the two waves.

Figure 4 Destructive interference. When two wave pulses that are out of phase come together, the resulting amplitude is reduced.

Technology Using Interference of Waves
Noise-cancelling headphones, shown in Figure 5, use the concept of destructive interference. The electronics inside the headphones generate a wave that is out of phase with sound waves in the exterior environment. This out-of-phase wave is played inside the headset. Using destructive interference, the outside noise is cancelled.
Such devices allow users to listen to music at lower volume levels, reducing potential damage to their hearing.

Figure 5 A detector inside the headset determines what noise there is, and a speaker in the headphone emits the out-of-phase wave.

## Free End Reflections Simulation

#### QUIZ

Q) When a wave pulse crosses the boundary from a fast medium into a slow medium

A. Both the transmitted and the reflected wave pulse are on the same side of the medium rest position
B. Both the transmitted and the reflected wave pulse are on the opposite side of the medium rest position
C. The transmitted wave pulse is on the same side, but the reflected wave pulse is on the opposite side
D. The reflected wave pulse is on the same side, but the transmitted wave pulse is on the opposite side

The transmitted wave pulse is on the same side, but the reflected wave pulse is on the opposite side

#### Free End Reflections Simulation

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Reflection
In our discussion of waves so far, we have assumed that the waves being analyzed could travel indefinitely without striking anything that would stop them or otherwise change their motion. But what happens to the motion of a wave when it reaches a boundary?

At a free boundary, waves are reflected
Consider a pulse wave traveling on a stretched rope whose end forms a ring around a post, as shown in Figure 01. We will assume that the ring is free to slide along the post without friction.
As the pulse travels to the right, each point of the rope moves up once and then back down. When the pulse reaches the boundary, the rope is free to move up as usual, and it pulls the ring up with it. Then, the ring is pulled back down by the tension in the rope. The movement of the rope at the post is similar to the movement that would result if someone were to whip the rope upward to send a pulse to the left, which would cause a pulse to travel back along the rope to the left. This is called reflection. Note that the reflected pulse is upright and has the same amplitude as the incident pulse.

Figure 01. When a wave in one medium (for example, string) encounters a medium with a lower density (for example, air), the wave is reflected with the same orientation and amplitude as the original pulse.

Media Boundaries: Amplitudes
The amplitude of a wave before it encounters a media boundary is closely related to the wave’s energy. The amplitude does not change if the wave’s energy remains constant.
When a wave encounters a media boundary that is not strictly an ideal free-end or fixed-end boundary, the wave splits into two. One wave is reflected, and the other is transmitted. The term transmission describes the process of a wave moving through a medium or moving from one medium into another medium. The amplitude of the original wave may not be shared equally by the reflected wave and the transmitted wave. However, the sum of the two amplitudes must equal the amplitude of the original wave (Figure 02).

Figure 02. At a media boundary that is neither free-end nor fixed-end, the original wave splits into two waves. (a) If the wave moving along the rope encounters a medium that has a slower wave speed, then the wave splits into two, and one wave is reflected and the other is transmitted. The reflected wave is upright. (b) When a wave moves into a faster medium, then the wave splits into two, and one wave is reflected and the other is transmitted. The reflected wave is inverted.

If the difference between the wave speeds in the two media is small, transmission is preferred—the amplitude of the transmitted wave is closer to the amplitude of the original wave. As a result, the amplitude of the reflected wave is much smaller because of the conservation of energy. For cases in which the wave speed is significantly different between the two media, reflection is preferred—the amplitude of the reflected wave is closer to the amplitude of the original wave.

## Fixed End Reflections Simulation

#### QUIZ

Q) When a wave pulse on a string reflects from a hard boundary, how is the reflected pulse related to the incident pulse?

A. Shape unchanged, amplitude unchanged
B. Shape inverted, amplitude unchanged
C. Shape unchanged, amplitude reduced
D. Shape inverted, amplitude reduced
E. Amplitude unchanged, speed reduced

#### Fixed End Reflections Simulation

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Reflection
In our discussion of waves so far, we have assumed that the waves being analyzed could travel indefinitely without striking anything that would stop them or otherwise change their motion. But what happens to the motion of a wave when it reaches a boundary?

Fixed End Reflections
If a medium is fixed at one end, then when a wave reaches the media boundary a fixed-end reflection occurs. A fixed-end reflection also occurs when a medium is fixed at both ends, as in a harp (Figure 01).
Consider a pulse in a string moving toward a rigid, denser medium such as a wall (Figure 01). When the pulse reaches the fixed end, it is reflected. As you see in Figure 01, the reflected pulse has the same shape as the incoming pulse, but its orientation is inverted. We may explain this inversion as follows. When the pulse reaches the fixed end of the string, it exerts an upward force on the wall. In response, the wall exerts a downward force on the string in accordance with Newton’s third law of motion.
Therefore, the incoming upright pulse is inverted upon reflection.

Figure 01. When a pulse in one medium meets a boundary with a denser medium, the reflected pulse is inverted.

The difference in the media as a wave reaches a media boundary may not be as dramatic as either the free-end or the fixed-end case. In nature there are media boundaries that are neither free-end nor fixed-end. For example, the boundary between water and air and the boundary between air and a tree are neither fixed-end boundaries nor free-end boundaries. If a wave travels from a medium in which its speed is faster to a medium in which its speed is slower, the wave particles can move more freely than they did in the faster medium. Th is may seem opposite to what you might think, but the motion of particles in media where the wave speed is higher is tightly controlled by molecular forces. So the transfer of the wave energy to the next particle is very efficient. Th is is similar to the free-end case, so the reflected wave has the same orientation as the original. Conversely, if a wave travels from a medium in which its speed is slower to a medium in which its speed is faster, the wave cannot move as freely. Th is is similar to the fixed-end case, so the reflected wave is inverted(see Figure 02 on the next).

Media Boundaries: Amplitudes
The amplitude of a wave before it encounters a media boundary is closely related to the wave’s energy. The amplitude does not change if the wave’s energy remains constant.
When a wave encounters a media boundary that is not strictly an ideal free-end or fixed-end boundary, the wave splits into two. One wave is reflected, and the other is transmitted. The term transmission describes the process of a wave moving through a medium or moving from one medium into another medium. The amplitude of the original wave may not be shared equally by the reflected wave and the transmitted wave. However, the sum of the two amplitudes must equal the amplitude of the original wave (Figure 02).

Figure 02. At a media boundary that is neither free-end nor fixed-end, the original wave splits into two waves. (a) If the wave moving along the rope encounters a medium that has a slower wave speed, then the wave splits into two, and one wave is reflected and the other is transmitted. The reflected wave is upright. (b) When a wave moves into a faster medium, then the wave splits into two, and one wave is reflected and the other is transmitted. The reflected wave is inverted.

If the difference between the wave speeds in the two media is small, transmission is preferred—the amplitude of the transmitted wave is closer to the amplitude of the original wave. As a result, the amplitude of the reflected wave is much smaller because of the conservation of energy. For cases in which the wave speed is significantly different between the two media, reflection is preferred—the amplitude of the reflected wave is closer to the amplitude of the original wave.

## Transverse Waves Simulation

#### Quiz

Q1) The distance between adjacent peaks in the direction of travel for a transverse wave is its

A. frequency.
B. period.
C. wavelength.
D. amplitude.

The wavelength of a transverse wave is also the distance between adjacent troughs, or between any adjacent identical parts of the waveform.

Q2) The vibrations along a transverse wave move in a direction

A. along the wave.
B. perpendicular to the wave.
C. Both A and B.
D. Neither A nor B.

The vibrations in a longitudinal wave, in contrast, are along (or parallel to) the direction of wave travel.

#### Transverse Wave Simulation

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Mechanical Waves
You have learned how Newton’s laws of motion and conservation of energy principles govern the behavior of particles. These laws also govern the motion of waves. There are many kinds of waves. All kinds of waves transmit energy, including the waves you cannot see, such as the sound waves you create when you speak and the light waves that reflect from the leaves on the trees.

Transverse waves
A wave is a rhythmic disturbance that carries energy through matter or space. Water waves, sound waves, and the waves that travel down a rope or spring are types of mechanical waves.
Mechanical waves require a medium. Water, air, ropes, or springs are the materials that carry the energy of mechanical waves. Other kinds of waves, including electromagnetic waves and matter waves, will be described in later chapters. Because many of these waves cannot be directly observed, mechanical waves can serve as models for their study.

FIGURE 14–1 A quick shake of a rope sends out wave pulses in both directions.

The two disturbances that go down the rope shown in Figure 14–1 are called wave pulses. A wave pulse is a single bump or disturbance that travels through a medium. If the person continues to move the rope up and down, a continuous wave is generated. Notice that the rope is disturbed in the vertical direction, but the pulse travels horizontally.
This wave motion is called a transverse wave. A transverse wave is a wave that vibrates perpendicular to the direction of wave motion.
FIGURE 14–4 These two photographs were taken 0.20 s apart. During that time, the crest moved 0.80 m. The velocity of the wave is 4.0 m/s.

Speed and amplitude
How fast does a wave move? The speed of the pulse shown in Figure 14–4 can be found in the same way in which you would determine the speed of a moving car. First, you measure the displacement of the wave peak, $\Delta d$; then you divide this by the time interval, $\Delta t$, to find the speed, as shown by $\Delta v = \Delta d/ \Delta t$. The speed of a continuous wave, can be found the same way. For most mechanical waves, both transverse and longitudinal, the speed depends only on the medium through which the waves move.
How does the pulse generated by gently shaking a rope differ from the pulse produced by a violent shake? The difference is similar to the difference between a ripple in a pond and a tidal wave. They have different amplitudes. The amplitude of a wave is its maximum displacement from its position of rest, or equilibrium. Two similar waves having different amplitudes are shown in Figure 14–5.

FIGURE 14–5 The amplitude of wave A is larger than that of wave B.

A wave’s amplitude depends on how the wave is generated, but not on its speed. More work has to be done to generate a wave with a larger amplitude. For example, strong winds produce larger water waves than those formed by gentle breezes.

FIGURE 14–6 One end of a string, with a piece of tape at point P, is attached to a vibrating blade. Note the change in position of point P over time.

Waves with larger amplitudes transfer more energy. Thus, although a small wave might move sand on a beach a few centimeters, a giant wave can uproot and move a tree. For waves that move at the same speed, the rate at which energy is transferred is proportional to the square of the amplitude. Thus, doubling the amplitude of a wave increases the amount of energy it transfers each second by a factor of four.

Wavelength
Rather than focusing on one point on a wave, imagine taking a snapshot of a wave, so that you can see the whole wave at one instant in time. Figure 14–5 shows the low points, or troughs, and the high points, or crests, of a wave. The shortest distance between points where the wave pattern repeats itself is called the wavelength. Crests are spaced by one wavelength. Each trough is also one wavelength from the next. The Greek letter lambda, $\lambda$ , represents wavelength.

Period and frequency
Although wave speed and amplitude can describe both pulses and continuous waves, period ($\ T$) and frequency ($\ f$) apply only to continuous waves. You learned in Chapter 6 that the period of a simple harmonic oscillator, such as a pendulum, is the time it takes for the motion of the oscillator to repeat itself. Such an oscillator is usually the source, or cause, of a continuous wave. The period of a wave is equal to the period of the source. In Figure 14–6, the period, $\ T$, equals 0.04 s, which is the time it takes the source to return to the same point in its oscillation. The same time is taken by point P, a point on the rope, to return to its initial position.
The frequency of a wave, $\ f$, is the number of complete oscillations it makes each second. Frequency is measured in hertz. One hertz (Hz) is one oscillation per second. The frequency ($\ f$) and period ($\ T$) of a wave are related by the following equation.

Frequency of a Wave $\ f = \frac{1}{T}$

Both the period and the frequency of a wave depend only on its source. They do not depend on the wave’s speed or the medium. Although you can measure a wavelength directly, the wavelength depends on both the frequency of the oscillator and the speed of the wave. In the time interval of one period, a wave moves one wavelength.
Therefore, the speed of a wave is the wavelength divided by the period, $\lambda /T$. Because the frequency is more easily found than the period, this equation is more often written as

Speed of a Wave  $\ v = \lambda f$.

## Rotational Equilibrium Simulation

#### Quiz

Q1) The net torque will be the same if a force is doubled but moved to a point halfway to the axis of rotation.

A. True
B. False

Torque is calculated by multiplying Force and distance

Q2) If the resultant of a set of concurrent forces is zero, the sum of the torques

A. is also zero
B. may or may not be zero
C. is not zero
D. depends on the axis of rotation

The sum of the moments of the forces about the concurrent point is zero, so the sum of the moments about every point is zero. The only condition imposed by equilibrium on a set of concurrent forces is that their sum is zero.

#### Rotational Equilibrium of seesaw Simulation

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Torques
Every time you open a door, turn on a water faucet, or tighten a nut with a wrench, you exert a turning force. These everyday movements are shown in Fig 01. Torque is produced by this turning force and tends to produce rotational acceleration. Torque is different from force. If you want to make an object move, apply a force. Unbalanced forces make things accelerate. To make an object turn or rotate, apply a torque. Torques produce rotation.

Fig 01. A torque produces rotation.

A doorknob is located as far as possible from the door’s hinge line for a good reason.
If you want to open a heavy door, you must certainly apply a force, but that is not enough. Where you apply that force and in what direction you push are also important. If you apply your force nearer to the hinge line than the knob, or at any angle other than 90 to the plane of the door, you must use a greater force than if you apply the force at the knob and perpendicular to the door’s plane.

As an example of what torque means and why it is important, think of the example of a door.
Fig 02. The torque is quite large when the force is applied far away from the fulcrum.

Fig 03. If the force is applied closer to the fulcrum, the torque is smaller.

Usually the door knob is placed far away from the fulcrum (where the door is attached to the wall with hinges) as shown in Fig 02. If you pull on the knob to open the door, you have a pretty easy time because the force. You are using causes a lot of torque(twisting) around the fulcrum (hinges) to open the door. Imagine that the door knob is now placed near the center of the door, closer to the hinge, as shown in Fig 03.
If you pull on the door knob with the same force as before, it will be more difficult to open the door. This is because the torque (twisting force) has decreased.

The formula to measure the amount of torque is a short one, but applying it depends on the question being asked.

$\ τ =r\times F$

$\ τ$ = torque (N∙m)
$\ F$ = force (N)
$\ r$ = lever arm distance (m)

At the end of a calculation of torque, it is common to give the torque a positive or negative value, depending on the direction it is turning.
◦ Clockwise (CW) torque = negative
◦ Counterclockwise (CCW) torque = positive
In the Figures shown above, the force was kept constant but the lever arm distance was changed. This is why we did not get as much torque in the second situation.

Fig 04. This scale relies on balanced torques.

Scale balances that work with sliding weights, such as the one shown in Fig 04, are based on balanced torques, not balanced masses. The sliding weights are adjusted until the counterclockwise torque just balances the clockwise torque. Then the arm remains horizontal. We say the scale is in rotational equilibrium.

Balanced Torques

Fig 05. A pair of torques can balance each other.

Torques are intuitively familiar to youngsters playing on a seesaw. Children can balance a seesaw even when their weights are not equal.
Weight alone does not produce a change in rotation—torque does. Children soon learn that the distance they sit from the pivot point is as important as their weight. In Fig 05, the person A sits a shorter distance from the fulcrum (turning axis) while the person B sits farther away. Balance is achieved if the torque that tends to produce clockwise rotation by the person A equals the torque that tends to produce counterclockwise rotation by the person B. When balanced torques act on an object, there is no change in rotation.

## Projectile motion 'monkey hunting' simulation

#### Quiz

Q) A hunter tries to shoot a rare monkey hanging from a tree with a dart gun. The hunter has the monkey in his sights. But the monkey notices the hunter and drops from the branch exactly when he hears the hunter blow. Will the hunter hit the monkey or miss the target?

A) The dart hit the monkey.
B) The dart miss above the monkey.
C) The dart miss below the monkey.

Monkey and the dart fell at the same rate, under the same influence of gravity. So the dart will hit the monkey.

#### Projectile motion - monkey hunting simulation

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A famous problem involving projectile motion is the “monkey and hunter problem” (Fig. 1). A hunter spots a monkey hanging from a tree branch, aims his rifle directly at the monkey, and fires. The monkey, hearing the shot, lets go of the branch at the same instant the hunter fires the rifle, hoping to escape by falling to the ground. Will the monkey escape? The unexpected answer is “no”: the bullet will always hit the monkey anyway, regardless of the angle of the rifle, the speed of the bullet, or the distance to the monkey, as long as the monkey is in range.

Fig 1. The “monkey and hunter” problem.

To show that this is so, let’s first define a coordinate system. Let the origin be at the end of the rifle, with the x axis pointing horizontally to the right, and $\ y$ pointing vertically upward. Let $\ D$ be the horizontal distance of the monkey from the origin, and $\ H$ be the initial height of the monkey (Fig. 1).
Now we’ll show that at $\ x=D$, the monkey and the bullet will both be at the same height $\ y$. Let the muzzle velocity of the rifle be $\ v_0$, and let the angle of fire of the rifle from the horizontal be . Let’s begin by finding the time $\ t$ needed for the bullet to travel the horizontal distance $\ D$ from the rifle. Since the horizontal component of the velocity is a constant $\ v0 cos\theta$, the $\ x$ coordinate of the bullet at time $\ t$ is

$\ x_{b} (t) = (v_0 cos\theta)t$ -- (1)

Setting $\ x_{b} (t) =D$ and solving for the time $\ t$, we find (calling this time $\ t_f$ )

$\ t_f = \frac{D}{v_0 cos \theta}$ -- (2)

Next, let’s find the $\ y$ coordinate of the bullet at this time $\ t_f$ .

By
$\ y_{b}(t) = -\frac{1}{2}gt^2 + (v_0 sin \theta)t$ -- (3)

Substituting $\ t = t_{f} = D/(v_0 cos\theta)$, we have

$\ y_b = -\frac{1}{2}g(\frac{D}{v_0cos\theta })^2 + (v_0 sin\theta) (\frac{D}{v_0cos\theta})$ -- (4)
= $\ -\frac{gD^2}{2v_0^2 cos^2 \theta} + D tan \theta$ -- (5)

But from trigonometry, $\ tan\theta = H/D$; making this substitution in the second term on the right, we have

$\ y_{b} = H - \frac{gD^2}{2v_0^2 cos^2 \theta}$ (bullet) -- (6)

Finally, let’s find the $\ y$ coordinate of the monkey at time $\ t_f$ . The monkey falls in one dimension; its $\ y$ coordinate at time $\ t$ is (using Eq. $\ y(t) = \frac{1}{2}at^2 + v_{0}t + y_{0}$  with $\ a= - g, v_{0}=0$ and $\ y_{0} = H$):

$\ y_{m} (t) = H - \frac{1}{2}gt^2$ -- (7)

Now substituting $\ t = t_{f} = D/(v_0 cos\theta)$,

$\ y_{m} = H - \frac{1}{2}g(\frac{D}{v_0cos\theta })^2$ -- (8)

$\ y_{m} = H - \frac{gD^2}{2v_0^2 cos^2 \theta}$ (monkey) -- (9)

Comparing Eqs. (9.53) and (9.56), you can see that the monkey and bullet will have the same $\ y$ coordinate when $\ x=D$, so the monkey will always get hit, regardless of the values of $\ D, H, v_{0}$, or $\ \theta.$

Essentially what’s happening here is that the monkey and bullet are both accelerated by the same amount, $\ g= 9.8 m/s^2$, so for a given amount of time, the monkey will fall the same distance as the bullet falls from the straight-line path it would take if there were no gravity. Therefore, the bullet always hits the monkey.

## Law of Reflection Simulation

Quiz

Quiz 1. Reflection is the process in which light strikes a surface and bounces off that surface. The reflected ray will bounce back directly to the light source if it is lined up with the ...

a. incident ray
b. reflected ray
c. normal line
d. reflecting surface

Quiz 2. When you attempt to focus an image on a screen, using a concave mirror, but cannot, yet, you can see an image when are looking into the same concave mirror, the image is called a ...

a. convex distortion
b. concave image
c. virtual image
d. reflected distortion

When a mirror cannot focus an image on a screen, but you can see an image in the mirror, it is called a virtual image

Law of Reflection Simulation

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REFLECTION OF LIGHT

Suppose you have just had your hair cut and you want to know what the back of your head looks like. You can do this seemingly impossible task by using two mirrors to direct light from behind your head to your eyes. Redirecting light with mirrors reveals a basic property of light’s interaction with matter.

Light traveling through a uniform substance, whether it is air, water, or a vacuum, always travels in a straight line. However, when the light encounters a different substance, its path will change. If a material is opaque to the light, such as the dark, highly polished surface of a wooden table, the light will not pass into the table more than a few wavelengths. Part of the light is absorbed, and the rest of it is deflected at the surface. This change in the direction of the light is called reflection. All substances absorb at least some incoming light and reflect the rest. A good mirror can reflect about 90 percent of the incident light, but no surface is a perfect reflector.

The texture of a surface affects how it reflects light
The manner in which light is reflected from a surface depends on the surface’s smoothness. Light that is reflected from a rough, textured surface, such as paper, cloth, or unpolished wood, is reflected in many different directions, as shown in Fig 1(a). This type of reflection is called diffuse reflection.
Light reflected from smooth, shiny surfaces, such as a mirror or water in a pond, is reflected in one direction only, as shown in Fig 1(b). This type of reflection is called specular reflection. A surface is considered smooth if its surface variations are small compared with the wavelength of the incoming light.
For our discussion, reflection will be used to mean only specular reflection.
(a)                                            (b)
Fig 1. Diffusely reflected light is reflected in many directions (a), whereas specularly reflected light is reflected in the same forward direction only (b).

(a)
(b)
Fig 2. The symmetry of reflected light (a) is described by the law of reflection, which states that the angles of the incoming and reflected rays are equal (b).

Incoming and reflected angles are equal
You probably have noticed that when incoming rays of light strike a smooth reflecting surface, such as a polished table or mirror, at an angle close to the surface, the reflected rays are also close to the surface.When the incoming rays are high above the reflecting surface, the reflected rays are also high above the surface. An example of this similarity between incoming and reflected rays is shown in Fig 2(a).
If a straight line is drawn perpendicular to the reflecting surface at the point where the incoming ray strikes the surface, the angle of incidence and the angle of reflection can be defined with respect to the line. Careful measurements of the incident and reflected angles q and q , respectively, reveal that the angles are equal, as illustrated in Fig 2(b).
(angle of incidence : the angle between a ray that strikes a surface and the line perpendicular to that surface at the point of contact
angle of reflection : the angle formed by the line perpendicular to a surface and the direction in which a reflected ray moves)

$\theta = \theta'$

angle of incoming light ray = angle of reflected light ray

The line perpendicular to the reflecting surface is referred to as the normal to the surface. It therefore follows that the angle between the incoming ray and the surface equals 90° − $\theta$, and the angle between the reflected ray and the surface equals 90° − $\theta'$ .