GRE Physics Companion: Doppler Effect and Moving Sources and Observers

GRE Physics Companion: Doppler Effect and Moving Sources and Observers

For one-dimensional acoustic propagation from source toward observer, take the propagation direction as positive.

Then

(1)

With this convention:

• v_s > 0 means the source moves toward the observer,
• v_o < 0 means the observer moves toward the source.

1 High-value GRE facts

1. Sound speed is measured relative to the medium.
2. A moving observer changes wavefront encounter rate.
3. A moving source changes wavelength.
4. With positive direction from source toward observer, f_o = f_s(c − v_o)∕(c − v_s).
5. Source motion toward the observer raises frequency.
6. Observer motion toward the source raises frequency.
7. At low speed, Δf∕f_s ≈ (v_s − v_o)∕c.
8. In multidimensional motion, use radial velocity components.
9. Reflection from a moving target produces a double Doppler shift.
10. The ordinary subsonic source formula breaks down at the sonic limit.

Part I: Original GRE-style problems

Problem 1: moving observer

A stationary source emits

(2)

sound in air with

(3)

An observer moves toward the source at

(4)

The observed frequency is closest to

1. 942 Hz
2. 1000 Hz
3. 1058 Hz
4. 1096 Hz
5. 1160 Hz

Problem 2: moving source

A source emitting

(5)

moves toward a stationary observer at

(6)

Take

(7)

The observed frequency is closest to

1. 920 Hz
2. 1000 Hz
3. 1058 Hz
4. 1096 Hz
5. 1160 Hz

Problem 3: source behind

For the source in Problem 2, a stationary observer behind the receding source hears a frequency closest to

1. 850 Hz
2. 920 Hz
3. 1000 Hz
4. 1096 Hz
5. 1180 Hz

Problem 4: both moving

A

(8)

source moves toward an observer at

(9)

while the observer moves toward the source at

(10)

For

(11)

the observed frequency is closest to

1. 920 Hz
2. 1058 Hz
3. 1096 Hz
4. 1160 Hz
5. 1300 Hz

Problem 5: low-speed shift

A source and observer approach one another with closing speed

(12)

For

(13)

the low-speed frequency shift is approximately

1. 0.5 Hz
2. 2.5 Hz
3. 5.0 Hz
4. 10 Hz
5. 50 Hz

Problem 6: radial motion

A source moves perpendicular to the line of sight at one instant. To first order in speed over

(14)

the instantaneous classical Doppler shift from that velocity component is

1. maximum positive
2. maximum negative
3. zero
4. infinite
5. independent of direction

Problem 7: wave speed

A sound source moves through still air. The speed of the emitted wavefronts relative to the air is

1. c − v_s
2. c
3. c + v_s
4. v_s
5. frequency dependent in the ideal nondispersive model

Problem 8: moving observer wavelength

A stationary source emits sound in still air. An observer begins moving toward the source. The wavelength in the air

1. decreases
2. increases
3. remains unchanged
4. becomes zero
5. depends on observer mass

Problem 9: reflected Doppler

For a stationary transmitter and receiver with a target approaching at speed

(15)

the exact classical reflected frequency is

1. f₀(c − u)∕(c + u)
2. f₀(c + u)∕(c − u)
3. f₀c∕(c − u)
4. f₀(c + u)∕c
5. f₀

Problem 10: double-shift approximation

For

(16)

the fractional reflected Doppler shift for an approaching target is approximately

1. u∕(2c)
2. u∕c
3. 2u∕c
4. c∕u
5. 2c∕u

Problem 11: moving medium

For acoustic Doppler calculations in moving air, source and observer velocities should be measured relative to

1. the ground only
2. the source only
3. the observer only
4. the air
5. an arbitrary inertial frame with no transformation

Problem 12: sonic limit

As a source moving toward a stationary observer approaches

(17)

from below, the simple subsonic forward-wave formula predicts an unbounded frequency. Physically this indicates

1. the wave speed becomes infinite
2. the source stops emitting
3. forward wavefronts pile up and the simple model breaks down
4. frequency becomes negative
5. observer motion must vanish

Part II: Complete worked solutions

Solution 1

Moving toward the source means

(18)

Thus

fo	= 1000	(19)
	≈ 1058.3 Hz.	(20)

Answer: (C).

Solution 2

fo	= 1000	(21)
	≈ 1095.8 Hz.	(22)

Answer: (D).

Solution 3

Behind the source,

fbehind	= 1000	(23)
	≈ 919.6 Hz.	(24)

Answer: (B).

Solution 4

Use

(25)

Then

fo	= 1000	(26)
	≈ 1159.7 Hz.	(27)

Answer: (D).

Solution 5

At low speed,

Δf	≈ fs	(28)
	= 500	(29)
	= 5.0 Hz.	(30)

Answer: (C).

Solution 6

Only the radial component contributes to the first-order shift. A perpendicular instantaneous velocity has zero radial component. Answer: (C).

Solution 7

In the ideal acoustic model, sound propagates through the air at speed

(31)

relative to the air. Answer: (B).

Solution 8

Observer motion changes encounter rate but does not alter the wavefront spacing already established in the air. Answer: (C).

Solution 9

The target experiences one Doppler shift while receiving and a second while returning the wave:

(32)

Answer: (B).

Solution 10

For

(33)

(34)

Answer: (C).

Solution 11

The medium defines the acoustic propagation frame. Velocities in the acoustic Doppler formula must therefore be relative to the air. Answer: (D).

Solution 12

As

(35)

forward wavefront spacing collapses. The divergence marks failure of the separated subsonic-wavefront construction. Answer: (C).

2 GRE checklist

1. Establish the propagation direction before choosing signs.
2. A moving observer changes encounter rate, not wavelength.
3. A moving source changes wavelength, not wave speed through the medium.
4. Use radial components in more than one dimension.
5. Use the exact classical formula before taking a low-speed limit.
6. Reflected Doppler involves two transformations.
7. Medium-relative velocities matter for sound.
8. Do not extend the subsonic formula through the sonic limit.

References

References