16.12

Concepts & Calculations

One of the important concepts that we encountered in this chapter is a transverse wave. For instance, transverse waves travel along a guitar string when it is plucked or along a violin string when it is bowed. The next example reviews how the travel speed depends on the properties of the string and on the tension in it.

| Concepts & Calcalations Example 12 | | | What Determines the Speed of a Wave on a String?

Figure 16-38 shows waves traveling on two strings. Each string is attached to a wall at one end and to a box that has a weight of 28.0 N at the other end. String 1 has a mass of 8.00 g and a length of 4.00 cm, and string 2 has a mass of 12.0 g and a length of 8.00 cm. Determine the speed of the wave on each string.

Figure 16-38 A wave travels on each of the two strings. The strings have different masses and lengths, and together support the 28.0-N box. Which is the faster wave?

Concept Questions and Answers Is the tension the same in each string?

Answer Yes, the tension is the same. The two strings support the box, so the tension in each string is one-half the weight of the box. The fact that the strings have different masses and lengths does not affect the tension, which is determined only by the weight of the hanging box.

Is the speed of each wave the same?

Answer Not necessarily. The speed of a wave on a string depends on both the tension and the linear density, as Equation 16.2 indicates. The tension is the same in both strings, but if the linear densities of the strings are different, the speeds are different.

String 1 has a smaller mass and, hence, less inertia than string 2. Does this mean that the speed of the wave on string 1 is greater than that on string 2?

Answer Maybe yes, maybe no. The speed of a wave depends on the linear density of the string, which is its mass divided by its length. Depending on the lengths of the strings, string 1 could have a larger linear density and, hence, smaller speed, than string 2. The solution below illustrates this point.

Solution The speed of a wave on a string is given by Equation 16.2 as , where F is the tension and m/L is the mass per unit length, or linear density. Since both strings support the box, the tension in each is one-half the weight of the box, or . The linear densities of the strings are

The speed of each wave is

The next example illustrates how the Doppler effect arises when an observer is moving away from or toward a stationary source of sound. In fact, we will see that it’s possible for both situations to occur at the same time.

| Concepts & Calcalations Example 13 | | | The Doppler Effect for a Moving Observer

A siren, mounted on a tower, emits a sound whose frequency is 2140 Hz. A person is driving a car away from the tower at a speed of 27.0 m/s. As Figure 16-39 illustrates, the sound reaches the person by two paths: the sound reflected from a building in front of the car, and the sound coming directly from the siren. The speed of sound is 343 m/s. What frequency does the person hear for the (a) reflected and (b) direct sounds?

Figure 16-39 The sound from the siren reaches the car by a reflected path and a direct path. The direct and reflected sound waves, as well as the motion of the car, are assumed to lie along the same line. Because of the Doppler effect, the driver hears a different frequency for each sound.

Concept Questions and Answers One way that the Doppler effect can arise is that the wavelength of the sound changes. For either the direct or the reflected sound, does the wavelength change?

Answer No. The wavelength changes only when the source of the sound is moving, as illustrated in Figure 16-29b. The siren is stationary, so the wavelength does not change.

Why does the driver hear a frequency for the reflected sound that is different than 2140 Hz, and is it greater than or smaller than 2140 Hz?

Answer The car and the reflected sound are traveling in opposite directions, the car to the right and the reflected sound to the left. The driver intercepts more wave cycles per second than if the car were stationary. Consequently, the driver hears a frequency greater than 2140 Hz.

Why does the driver hear a frequency for the direct sound that is different than 2140 Hz, and is it greater than or smaller than 2140 Hz?

Answer The car and the direct sound are traveling in the same direction. As the direct sound passes the car, the number of wave cycles per second intercepted by the driver is less than if the car were stationary. Thus, the driver hears a frequency that is less than 2140 Hz.

Solution

a. | For the reflected sound, the frequency fo that the driver (the “observer”) hears is equal to the frequency fs of the waves emitted by the siren plus an additional number of cycles per second because the car and the reflected sound are moving in opposite directions. The additional number of cycles per second is , where vo is the speed of the car and is the wavelength of the sound (see the subsection “Moving Observer” in Section 16.9). According to Equation 16.1, the wavelength is equal to the speed of sound v divided by the frequency of the siren, . Thus, the frequency heard by the driver can be written as

b. | For the direct sound, the frequency fo that the driver hears is equal to the frequency fs of the waves emitted by the siren minus , because the car and direct sound are moving in the same direction:

As expected, for the reflected wave, the driver hears a frequency greater than 2140 Hz, while for the direct sound he hears a frequency less than 2140 Hz.

At the end of the problem set for this chapter, you will find homework problems that contain both conceptual and quantitative parts. These problems are grouped under the heading Concepts & Calculations.

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