Waves

1

A police car is at a red light. A car behind him foolishly tests his luck and runs the red light. The officer immediately flicks on his siren and begins to chase the car, accelerating at a rate of $1.5\frac{m}{s^2}$ . The frequency of the siren is 500Hz. If you are standing at the red light, what frequency do you percieve 6 seconds after the police officer began chasing the car?

$v_s=340\frac{m}{s}$

Explanation

We simply need to know the Doppler effect equation to solve this problem:

$f' = \left ( \frac{v\pm v_o}{v\pm v_s}\right )f_o$

Since the observer (you) is not moving, we can rewrite the equation as:

$f' = \left ( \frac{v}{v+ v_s}\right )f_o$

The reason for using addition in the denominator is explained below.

We know all of our values, so we can simply solve for the perceived frequency:

$f' = \left ( \frac{340}{340+9} \right )500 = 487 hz$

For the Doppler effect equation, don't waste your time trying to memorize whether the signs should be $\pm\ \text{or}\ \mp$ . We have purposely made them the same to emphasize the following idea. Think about the situation practically. The police car is moving away from you. Therefore, we know that the percieved frequency is going to be less than the source. How do we make the frequency lower? We either lower the numerator or increase the denominator. Since we can only manipulate the denominator, we will use the + sign. Knowing this will also help you immediately eliminate all answers in which the frequency is increased.

2

A person showing off their new sound system is driving towards a group of people at a speed of $60 \frac{km}{h}$ . The group of people is stationary. What is the observed frequency of the sound if the sound emitted by the car has a frequency of 421Hz?

$v_{sound}=343\frac{m}{s}$

Explanation

The difference between the observed frequency and the frequency of the source due to relative velocity is called the Doppler effect, and is given by the following formula:

$f_{o}=f_{s}(\frac{v\pm v_{0}}{v\pm v_{s}})$

where $f_{o}$ is the observed frequency, $f_{s}$ is the source frequency, is the speed of sound, $v_{o}$ is the speed of the observer, and $v_{s}$ is the speed of the source. The speed of the observer is equal to zero, because the group of people is stationary. The speed of sound was given to us as $343\frac{m}{s}$ . The frequency of the source was given as 421Hz. We must convert the speed of the source into $\frac{m}{s}$ , which when converted is equal to $16.67\frac{m}{s}$ . Finally, we must determine whether the sign in the denominator term should be a plus or minus; because we are approaching the group of people, the observed frequency should be higher, therefore a minus sign is appropriate (makes a smaller denominator which makes a larger result). Our final answer after plugging everything in is 442Hz.

$f_o=421*\frac{343+0}{343-16.67}=442Hz$

3

You are driving your car by a very loud concert and moving at $13 \frac{m}{s}$ . If the frequency of a particular long note is 740 Hz, what is the frequency of the note you hear as you approach the concert? What is the frequency of the note you hear as you move away from the concert? The speed of sound in air is $343\frac{m}{s}$ .

Moving toward the concert:

Moving away from the concert:

Moving toward the concert:

Moving away from the concert:

Moving toward the concert:

Moving away from the concert:

Moving toward the concert:

Moving away from the concert:

Explanation

Remember the equation for the doppler effect for a moving observer:

$f'=(1\pm \frac{u}{v})f$

Now, identify the given information:

$u=13\frac{m}{s}$

$v=343\frac{m}{s}$ (This is the speed of sound in air.)

When you are moving towards the concert, the plus sign is used. Therefore, the apparent frequency is

$f'=(1+\frac{13\frac{m}{s}}{343\frac{m}{s}})(740Hz)=768Hz$

When you are moving away from the concert, the negative sign is used. Therefore, the apparent frequency is

$f'=(1-\frac{13\frac{m}{s}}{343\frac{m}{s}})(740Hz)=712Hz$

4

A police car is at a red light. A car behind him foolishly tests his luck and runs the red light. The officer immediately flicks on his siren and begins to chase the car, accelerating at a rate of $1.5\frac{m}{s^2}$ . The frequency of the siren is 500Hz. If you are standing at the red light, what frequency do you percieve 6 seconds after the police officer began chasing the car?

$v_s=340\frac{m}{s}$

Explanation

We simply need to know the Doppler effect equation to solve this problem:

$f' = \left ( \frac{v\pm v_o}{v\pm v_s}\right )f_o$

Since the observer (you) is not moving, we can rewrite the equation as:

$f' = \left ( \frac{v}{v+ v_s}\right )f_o$

The reason for using addition in the denominator is explained below.

We know all of our values, so we can simply solve for the perceived frequency:

$f' = \left ( \frac{340}{340+9} \right )500 = 487 hz$

For the Doppler effect equation, don't waste your time trying to memorize whether the signs should be $\pm\ \text{or}\ \mp$ . We have purposely made them the same to emphasize the following idea. Think about the situation practically. The police car is moving away from you. Therefore, we know that the percieved frequency is going to be less than the source. How do we make the frequency lower? We either lower the numerator or increase the denominator. Since we can only manipulate the denominator, we will use the + sign. Knowing this will also help you immediately eliminate all answers in which the frequency is increased.

5

A person showing off their new sound system is driving towards a group of people at a speed of $60 \frac{km}{h}$ . The group of people is stationary. What is the observed frequency of the sound if the sound emitted by the car has a frequency of 421Hz?

$v_{sound}=343\frac{m}{s}$

Explanation

The difference between the observed frequency and the frequency of the source due to relative velocity is called the Doppler effect, and is given by the following formula:

$f_{o}=f_{s}(\frac{v\pm v_{0}}{v\pm v_{s}})$

where $f_{o}$ is the observed frequency, $f_{s}$ is the source frequency, is the speed of sound, $v_{o}$ is the speed of the observer, and $v_{s}$ is the speed of the source. The speed of the observer is equal to zero, because the group of people is stationary. The speed of sound was given to us as $343\frac{m}{s}$ . The frequency of the source was given as 421Hz. We must convert the speed of the source into $\frac{m}{s}$ , which when converted is equal to $16.67\frac{m}{s}$ . Finally, we must determine whether the sign in the denominator term should be a plus or minus; because we are approaching the group of people, the observed frequency should be higher, therefore a minus sign is appropriate (makes a smaller denominator which makes a larger result). Our final answer after plugging everything in is 442Hz.

$f_o=421*\frac{343+0}{343-16.67}=442Hz$

6

You are driving your car by a very loud concert and moving at $13 \frac{m}{s}$ . If the frequency of a particular long note is 740 Hz, what is the frequency of the note you hear as you approach the concert? What is the frequency of the note you hear as you move away from the concert? The speed of sound in air is $343\frac{m}{s}$ .

Moving toward the concert:

Moving away from the concert:

Moving toward the concert:

Moving away from the concert:

Moving toward the concert:

Moving away from the concert:

Moving toward the concert:

Moving away from the concert:

Explanation

Remember the equation for the doppler effect for a moving observer:

$f'=(1\pm \frac{u}{v})f$

Now, identify the given information:

$u=13\frac{m}{s}$

$v=343\frac{m}{s}$ (This is the speed of sound in air.)

When you are moving towards the concert, the plus sign is used. Therefore, the apparent frequency is

$f'=(1+\frac{13\frac{m}{s}}{343\frac{m}{s}})(740Hz)=768Hz$

When you are moving away from the concert, the negative sign is used. Therefore, the apparent frequency is

$f'=(1-\frac{13\frac{m}{s}}{343\frac{m}{s}})(740Hz)=712Hz$

7

A pendulum of length with a ball of mass is released at an angle $\theta$ away from the equilibrium point. Which of the following adjustments will result in an increase in frequency of oscillation?

Decreasing the length

Increasing the ball's mass

Decreasing the ball's mass

Increasing the length

Increasing the initial angle $\theta$

Explanation

For a pendulum on a string, the period at which it oscillates is:

$T = 2\pi \sqrt{\frac{l}{g}}$

Period is the reciprocal of frequency.

$T = \frac{1}{f}$

Calculate the frequency of oscillation.

$f = \frac{1}{2\pi}\sqrt{\frac{g}{l}}$

The only parameter that the frequency depends on is the length . Decreasing length increases frequency.

8

A pendulum of length with a ball of mass is released at an angle $\theta$ away from the equilibrium point. Which of the following adjustments will result in an increase in frequency of oscillation?

Decreasing the length

Increasing the ball's mass

Decreasing the ball's mass

Increasing the length

Increasing the initial angle $\theta$

Explanation

For a pendulum on a string, the period at which it oscillates is:

$T = 2\pi \sqrt{\frac{l}{g}}$

Period is the reciprocal of frequency.

$T = \frac{1}{f}$

Calculate the frequency of oscillation.

$f = \frac{1}{2\pi}\sqrt{\frac{g}{l}}$

The only parameter that the frequency depends on is the length . Decreasing length increases frequency.

9

What is the apparent frequency of a wave traveling at a speed and frequency that is moving away from you at a speed , and that you are moving towards at a speed ? Note: in this context is not referring to the speed of light.

$f'=f\left(\frac{v+ 2c}{v+ c}\right)$

$f'=f\left(\frac{v+ 2c}{v- c}\right)$

Explanation

The formula for the doppler effect is:

$f'=f\left(\frac{v\pm v_D}{v\mp v_s}\right)$ , where is the apparent frequency of the wave, is the actual frequency, is the wave velocity, is the velocity of the detector, is the velocity of the source. In our case, the detector (which is you) is moving towards the source at speed , making the numerator a plus sign. The source is moving away from us at speed , making the denominator also a plus sign. Knowing this, we can now plug into our equation:

$f'=f\left(\frac{v+ 2c}{v+ c}\right)$

10

What is the apparent frequency of a wave traveling at a speed and frequency that is moving away from you at a speed , and that you are moving towards at a speed ? Note: in this context is not referring to the speed of light.

$f'=f\left(\frac{v+ 2c}{v+ c}\right)$

$f'=f\left(\frac{v+ 2c}{v- c}\right)$

Explanation

The formula for the doppler effect is:

$f'=f\left(\frac{v\pm v_D}{v\mp v_s}\right)$ , where is the apparent frequency of the wave, is the actual frequency, is the wave velocity, is the velocity of the detector, is the velocity of the source. In our case, the detector (which is you) is moving towards the source at speed , making the numerator a plus sign. The source is moving away from us at speed , making the denominator also a plus sign. Knowing this, we can now plug into our equation:

$f'=f\left(\frac{v+ 2c}{v+ c}\right)$

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