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Chapter 16: Problem 80

The security alarm on a parked car goes off and produces a frequency of 960 Hz. The speed of sound is 343 m/s. As you drive toward this parked car, pass it, and drive away, you observe the frequency to change by 95 Hz. At what speed are you driving?

Short Answer

Expert verified
The driving speed is approximately 17.0 m/s.

Step by step solution

01

Understanding the Doppler Effect

The Doppler Effect describes how the frequency of a sound changes with the relative motion between the sound source and an observer. When moving toward the source, the observed frequency increases; when moving away, it decreases.
02

Define the Variables

Given: \( f_0 = 960 \text{ Hz} \) is the source frequency, \( v_s = 343 \text{ m/s} \) is the speed of sound, and \( \Delta f = 95 \text{ Hz} \) is the observed change in frequency. We are to determine \( v_o \), the speed of the driver.
03

Apply the Doppler Effect Formula for Approaching and Receding Sounds

The Doppler Effect formula for an observer moving relative to a stationary source is \( f = \frac{v_s + v_o}{v_s} f_0 \) when approaching, and \( f = \frac{v_s - v_o}{v_s} f_0 \) when receding. The frequency changes from \( f_1 \) to \( f_2 \), and the difference \( \Delta f = f_1 - f_2 \) is given as 95 Hz.
04

Calculate Observed Frequencies

Let \( f_1 = \frac{v_s + v_o}{v_s} f_0 \) and \( f_2 = \frac{v_s - v_o}{v_s} f_0 \). Then, the change in frequency, \( \Delta f = f_1 - f_2 = 95 \text{ Hz} \), can be expressed as: \( \frac{v_s + v_o}{v_s} f_0 - \frac{v_s - v_o}{v_s} f_0 = 95 \).
05

Simplify and Solve for Velocity

Combine the expressions: \( \left( \frac{v_s + v_o}{v_s} - \frac{v_s - v_o}{v_s} \right) \times 960 = 95 \). Simplify to get: \( \frac{2v_o}{v_s} \times 960 = 95 \). Solving for \( v_o \): \( 2v_o \times 960 = 95 \times 343 \). Thus, \( v_o = \frac{95 \times 343}{2 \times 960} \approx 17.0 \text{ m/s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sound Frequency
Sound frequency refers to the number of vibrations or waves produced per second by a sound source. The unit of frequency is hertz (Hz). In our example, the car alarm emits a frequency of 960 Hz; this means the sound wave completes 960 cycles every second.
Sound frequencies can vary depending on the source and are a key aspect in understanding sound perception and phenomena like the Doppler Effect.
  • Higher frequencies result in higher-pitched sounds.
  • Lower frequencies result in lower-pitched sounds.
The frequency of the sound a driver hears may change when the driver or the sound source is in motion, as we will explore further in the section about observed frequency change.
Speed of Sound
The speed of sound is the rate at which sound waves travel through a medium. In air at room temperature, this speed is approximately 343 meters per second (m/s). The speed of sound can change based on factors like medium, temperature, and pressure.
Understanding the speed of sound is crucial when studying how sound waves interact with moving objects. In the given problem, this constant helps us determine how the sound frequency changes when the listener is moving.
  • The speed of sound is faster in solids compared to liquids and gases due to the density and elasticity of the medium.
  • Changes in temperature can slightly increase or decrease the speed of sound in the air.
Knowing the speed of sound is vital in calculating the changes in observed frequency when considering the Doppler Effect.
Observed Frequency Change
The observed frequency change occurs when there is a relative motion between the sound source and the observer. This is described by the Doppler Effect. In our scenario, as the car alarm produces a frequency of 960 Hz, the frequency that the driver hears changes as they move closer to and further from the car.
The exercise shows that the driver's motion results in a frequency change of 95 Hz, first increasing as they approach and decreasing as they drive away.
  • Approaching the source increases the observed frequency.
  • Moving away decreases the observed frequency.
These changes in frequency occur because sound waves are compressed when the observer moves toward the source and stretched when moving away. Calculating these variations requires both the original frequency and the speed of sound, revealing insights into the speed at which the observer, in this case, the driver, is moving.

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Most popular questions from this chapter

An explosion occurs at the end of a pier. The sound reaches the other end of the pier by traveling through three media: air, fresh water, and a slender metal handrail. The speeds of sound in air, water, and the handrail are 343, 1482, and 5040 m/s, respectively. The sound travels a distance of 125 m in each medium. (a) Through which medium does the sound arrive first, second, and third? (b) After the first sound arrives, how much later do the second and third sounds arrive?

To navigate, a porpoise emits a sound wave that has a wavelength of 1.5 cm. The speed at which the wave travels in seawater is 1522 m/s. Find the period of the wave.

The speed of a sound in a container of hydrogen at 201 K is 1220 m/s. What would be the speed of sound if the temperature were raised to 405 K? Assume that hydrogen behaves like an ideal gas.

mm At what temperature is the speed of sound in helium (ideal gas, \(\gamma=1.67,\) atomic mass \(=4.003\) u) the same as its speed in oxygen at \(0^{\circ} \mathrm{C} ?\)

An ultrasonic ruler, such as the one discussed in Example 4 in Section 16.6, displays the distance between the ruler and an object, such as a wall. The ruler sends out a pulse of ultrasonic sound and measures the time it takes for the pulse to reflect from the object and return. The ruler uses this time, along with a preset value for the speed of sound in air, to determine the distance. Suppose that you use this ruler under water, rather than in air. The actual distance from the ultrasonic ruler to an object is 25.0 m. The adiabatic bulk modulus and density of seawater are \(B_{\mathrm{ad}}=2.37 \times 10^{9} \mathrm{Pa}\) and \(\rho=1025 \mathrm{kg} / \mathrm{m}^{3}\) , respectively. Assume that the ruler uses a preset value of 343 \(\mathrm{m} / \mathrm{s}\) for the speed of sound in air. Determine the distance reading that the ruler displays.
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