Question 1

A stationary source generates 5.0 Hz water waves whose speed is 2.0 m/s.A boat is approaching the source at 1.0 m/s.The frequency of these waves, as observed by a person in the boat, is:
A)2.5 Hz
B)5.0 Hz
C)7.5 Hz
D)15 Hz
E)30 Hz

Accepted Answer

The frequency observed by a person in the boat can be calculated using the formula:

f' = (v +/- v_r)/lambda

where f' is the observed frequency, v is the speed of the wave, v_r is the velocity of the observer relative to the source, and lambda is the wavelength.

In this case, the frequency of the wave generated by the source is 5.0 Hz, and the speed of the wave is 2.0 m/s. The boat is approaching the source at 1.0 m/s, so the velocity of the observer relative to the source is -1.0 m/s (negative because the boat is moving towards the source).

The wavelength of the wave can be calculated using the formula:

lambda = v/f

where f is the frequency of the wave.

Substituting the given values, we get:

lambda = 2.0/5.0 = 0.4 m

Substituting these values in the formula for observed frequency, we get:

f' = (2.0 - 1.0)/0.4 = 2.5 Hz (using the minus sign because the observer is approaching the source)

Therefore, the frequency of the waves observed by a person in the boat is 2.5 Hz, which corresponds to choice C.

Question 2

A source emits sound with a frequency of 1000 Hz.It and an observer are moving toward each other, each with a speed of 100 m/s.If the speed of sound is 340 m/s, the observer hears sound with a frequency of:
A)290 Hz
B)540 Hz
C)1000 Hz
D)1800 Hz
E)3400 Hz

Accepted Answer

The observed frequency can be calculated using the formula for the Doppler effect when both the source and the observer are moving towards each other: $f' = f \frac{v + v_o}{v - v_s}$, where $f'$ is the observed frequency, $f$ is the source frequency, $v$ is the speed of sound, $v_o$ is the speed of the observer towards the source, and $v_s$ is the speed of the source towards the observer. Plugging in the given values: $f' = 1000 \frac{340 + 100}{340 - 100} = 1000 \frac{440}{240} = 1000 \times \frac{11}{6} = 1833.33 $ Hz, which rounds to 1800 Hz, option D.

Question 3

A source emits sound with a frequency of 1000 Hz.It is moving at 20 m/s toward a stationary reflecting wall.If the speed of sound is 340 m/s an observer at rest directly behind the source hears a beat frequency of:
A)3.0 Hz
B)55 Hz
C)63 Hz
D)114 Hz
E)118 Hz

Accepted Answer

The answer of A source emits sound with a frequency...

Question 4

A stationary source S generates circular outgoing waves on a lake.The wave speed is 5.0 m/s and the crest-to-crest distance is 2.0 m.A person in a motor boat heads directly toward S at 3.0 m/s.To this person, the frequency of these waves is:
A)1.0 Hz
B)1.5 Hz
C)2.0 Hz
D)4.0 Hz
E)8.0 Hz

Accepted Answer

The answer of A stationary source S generates circular outgoing...

Question 5

A stationary source emits a sound wave of frequency f.If it were possible for a man to travel toward the source at the speed of sound, he would observe the emitted sound to have a frequency of:
A)0
B)f/2
C)2f/3
D)2f
E)infinity

Accepted Answer

This is an application of the Doppler effect. When an observer moves towards a stationary source, the effective frequency of the wave increases. In this case, the observer is moving towards the source at the speed of sound, so the relative velocity between the observer and the source is the speed of sound. The formula for the Doppler effect in this case is:

f' = f (v_sound + v_observer) / (v_sound + v_source)

where f is the frequency of the wave emitted by the source, f' is the frequency observed by the observer, v_sound is the velocity of sound, and v_observer and v_source are the velocities of the observer and the source, respectively.

Since the observer is moving towards the source at the speed of sound, v_observer = v_sound. Therefore:

f' = f (v_sound + v_sound) / (v_sound + v_source) = 2f(v_sound) / (v_sound + v_source)

As the observer approaches the source, v_source becomes less than v_sound, so the denominator of this fraction decreases, and the effective frequency observed by the observer increases. When v_source = 0 (i.e. the source is stationary), we get:

f' = 2f(v_sound) / (v_sound + 0) = 2f

Therefore, the observed frequency would be twice the emitted frequency. The correct choice is D.

Question 6

The diagram shows four situations in which a source of sound S and a detector D are either moving or stationary.The arrows indicate the directions of motion.The speeds (when not zero)are all the same.(Note that the detector in situation 3 is stationary).Rank the situations according to the apparent frequency of the source, lowest to highest. 
A)1, 2, 3, 4
B)4, 3, 2, 1
C)1, 3, 4, 2
D)2, 1, 4, 3
E)None of the above

Accepted Answer

The answer of The diagram shows four situations in which...

Question 7

A source emits sound with a frequency of 1000 Hz.Both it and an observer are moving in the same direction with the same speed, 100 m/s.If the speed of sound is 340 m/s, the observer hears sound with a frequency of:
A)290 Hz
B)540 Hz
C)1000 Hz
D)1800 Hz
E)3400 Hz

Accepted Answer

Since both the source and the observer are moving at the same speed in the same direction, there is no relative motion between them. Therefore, the observer hears the sound at its original frequency, which is 1000 Hz.

Question 8

The speed of sound is 340 m/s.A plane flies horizontally at an altitude of 10,000 m and a speed of 400 m/s.When an observer on the ground hears the sonic boom the horizontal distance from the point on its path directly above the observer to the plane is (assume the speed of sound does not change with altitude):
A)5800 m
B)6100 m
C)8400 m
D)12,000 m
E)16,000 m

Accepted Answer

The time taken by the sound wave to reach the observer on the ground is given by the distance traveled by the sound wave divided by the speed of sound. Let the horizontal distance from the point on the plane's path directly above the observer to the observer be x. Then, using the Pythagorean theorem, the total distance traveled by the sound wave is given by:

distance = sqrt(x^2 + 10000^2)

The time taken for the plane to travel this distance is given by:

time = distance / plane speed = [sqrt(x^2 + 10000^2)] / 400

The time taken by the sound wave to reach the observer is given by:

time = distance / speed of sound = [sqrt(x^2 + 10000^2)] / 340

Since the observer hears the sonic boom at the same time the plane passes overhead, we can equate the two expressions for time and solve for x:

[sqrt(x^2 + 10000^2)] / 400 = [sqrt(x^2 + 10000^2)] / 340

Simplifying this equation gives:

6x = 17,000

x = 2833.33 meters

Therefore, the horizontal distance from the point on the plane's path directly above the observer to the plane is approximately 6100 meters (B).

Question 9

A plane produces a sonic boom only when:
A)its speed changes from being slower than the speed of sound to being faster than the speed of sound
B)it emits sound waves of high frequency
C)it flies at high altitudes
D)it flies on a curved path
E)it flies faster than the speed of sound

Accepted Answer

A sonic boom is a loud sound created by shock waves formed due to the disturbances caused by an object traveling through the air faster than the speed of sound. Therefore, a plane must fly faster than the speed of sound to produce a sonic boom.

Question 10

If the speed of sound is 340 m/s a plane flying at 400 m/s has a Mach number of:
A)0.85
B)1.2
C)1.4
D)1.7
E)400

Accepted Answer

The Mach number is calculated by dividing the object's speed by the speed of sound in the same medium. Here, Mach number = 400 m/s / 340 m/s = 1.176, which rounds to 1.2.

Question 11

If the speed of sound is 340 m/s a plane flying at 400 m/s creates a conical shock wave with an apex half angle of:
A)"0° (no shock wave)
B)"32 ° "
C)"40 ° "
D)"50 ° "
E)"58 ° "

Accepted Answer

The Mach angle is calculated using the formula sin(θ) = 1/M, where M is the Mach number. The Mach number is the speed of the plane divided by the speed of sound, which is 400/340 ≈ 1.176. Therefore, θ = sin⁻¹(1/1.176) ≈ 58°.

Question 12

In each of the following two situations a source emits sound with a frequency of 1000 Hz.In situation I the source is moving at 100 m/s toward an observer at rest.In situation II the observer is moving at 100 m/s toward the source, which is stationary.The speed of sound is 340 m/s.The frequencies heard by the observers in the two situations are:
A)I: 1417 Hz; II: 1294 Hz
B)I: 1417 Hz; II: 1417 Hz
C)I: 1294 Hz; II: 1294 Hz
D)I: 773 Hz; II: 706 Hz
E)I: 773 Hz; II: 773 Hz

Accepted Answer

In situation I, the source is moving towards the observer. This means that the frequency of the sound waves will appear higher to the observer. We can use the formula:

$f_{obs}=\left(\frac{v\pm v_{obs}}{v\pm v_s}\right)f_s$

Where $f_s$ is the frequency emitted by the source, $v$ is the speed of sound, $v_s$ is the velocity of the source (positive when moving towards the observer), and $v_{obs}$ is the velocity of the observer (positive when moving towards the source).

Plugging in the values for situation I, we get:

$f_{obs}=\left(\frac{340+100}{340}\right)1000=1417 Hz$

In situation II, the observer is moving towards the source. This means that the frequency of the sound waves will also appear higher to the observer. Plugging in the values for situation II, we get:

$f_{obs}=\left(\frac{340}{340-100}\right)1000=1294 Hz$

Therefore, the answer is choice A) I: 1417 Hz; II: 1294 Hz.

Quiz 17: Waves

A stationary source generates 5.0 Hz water waves whose speed is 2.0 m/s.A boat is approaching the source at 1.0 m/s.The frequency of these waves, as observed by a person in the boat, is:

A source emits sound with a frequency of 1000 Hz.It and an observer are moving toward each other, each with a speed of 100 m/s.If the speed of sound is 340 m/s, the observer hears sound with a frequency of:

A source emits sound with a frequency of 1000 Hz.It is moving at 20 m/s toward a stationary reflecting wall.If the speed of sound is 340 m/s an observer at rest directly behind the source hears a beat frequency of:

A stationary source S generates circular outgoing waves on a lake.The wave speed is 5.0 m/s and the crest-to-crest distance is 2.0 m.A person in a motor boat heads directly toward S at 3.0 m/s.To this person, the frequency of these waves is:

A stationary source emits a sound wave of frequency f.If it were possible for a man to travel toward the source at the speed of sound, he would observe the emitted sound to have a frequency of:

A source emits sound with a frequency of 1000 Hz.Both it and an observer are moving in the same direction with the same speed, 100 m/s.If the speed of sound is 340 m/s, the observer hears sound with a frequency of:

A plane produces a sonic boom only when:

If the speed of sound is 340 m/s a plane flying at 400 m/s has a Mach number of:

If the speed of sound is 340 m/s a plane flying at 400 m/s creates a conical shock wave with an apex half angle of: