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AP Physics 1 : Sound Waves

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Example Questions

Example Question #1 : Sound Waves

Alice measures the wavelength and frequency of a sound wave as

$\lambda=3.5m$

T=0.01s

At what speed is the sound traveling?

Possible Answers:

$28.6 \frac{m}{s}$

$3.5\frac{m}{s}$

$427\frac{m}{s}$

$350 \frac{m}{s}$

$0.00286\frac{m}{s}$

Correct answer:

$350 \frac{m}{s}$

Explanation:

We know from the question that

$\lambda=3.5m$

T=0.01s

Frequency is the inverse of period:

$f=\frac{1}{T}=\frac{1}{0.01s}=100s^{-1}}$

The velocity of a wave is its wavelength multiplied by its frequency:

$v=\lambda f = (3.5 m)(100s^{-1})=350\frac{m}{s}$

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Example Question #2 : Sound Waves

You are standing on the sidewalk when a police car approaches you at $30\frac{m}{s}$ with its sirens on. Its sirens seem to have a frequency of 500 Hertz. After the police car passes you and is driving away, what will be the new frequency you hear?

$v_{sound\:in\:air}=340\frac{m}{s}$

Possible Answers:

544Hz

419 Hz

507Hz

456Hz

500Hz

Correct answer:

419 Hz

Explanation:

The doppler effect follows this formula:

$f=\frac{s}{s+v}f_0$

In this equation, is the new frequency you will hear, is the speed of sound, is the velocity of the moving sound-emitting thing, and f_0 is the initial frequency of the sound.

Plugging the given values in, we can describe the initial situation as:

$500=\frac{340}{340-30}f_0$

Note that the velocity is negative (-30) because the car is driving towards you.

Therefore,

$f_0=456s^{-1}$

When the police car is driving away, the situation is described with a positive velocity:

$f=\frac{340}{340+30}456$

Therefore,

$f=419s^{-1}$

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Example Question #1 : Sound Waves

A guitar player uses beats to tune his instrument by playing two strings. If one vibrates at 550 Hz and the second at 555 Hz, how many beats will he hear per minute?

Possible Answers:

$\frac{1}{12} \textup{\textup{ bpm}}$

$5 \textup{\textup{ bpm}}$

$50 \textup{\textup{ bpm}}$

$300 \textup{\textup{ bpm}}$

$\frac{1}{2} \textup{\textup{ bpm}}$

Correct answer:

$300 \textup{\textup{ bpm}}$

Explanation:

Two waves will emit a beat with a fequency equal to the difference in frequency of the two waves. In this case the beat frequency is:

$555 - 550=5 \textup{\textup{ Hz}}$ (beats per second)

Convert to beats per minute:

$5 \textup{\textup{ Hz}}\times \frac{60}{1} \frac{seconds}{minutes} = 300 \textup{\textup{ bpm}}$

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Example Question #1 : Sound Waves

The ukulele is a short instrument, relative to a guitar. How does this affect the frequencies of sounds that these two instruments produce? Assume the two instruments use the same strings.

Possible Answers:

The shorter length creates a lower speed of sound

The shorter length strings produce higher frequencies

The shorter length creates a higher speed of sound

The shorter length strings produce lower frequencies

Correct answer:

The shorter length strings produce higher frequencies

Explanation:

The speed of sound in air is constant, assuming that the temperature of the air is constant. When the length of the string is shortened, by the principles of standing waves, this creates a higher frequencies. Assuming that the two instruments use the same strings is equivalent to stating that the two instruments have strings of equal linear mass density. This situation represents a standing wave, thus we can relate the following equation for the first harmonic:

$L=2\lambda$

Where, is the length of the string and $\lambda$ is the wavelength. Then we can use the following equation to relate wavelength and speed (which is known) to frequency:

$v=\lambda f$

Since the velocity of sound in a fixed medium is constant, we see that a shorter length, corresponds to a shorter wavelength, $\lambda$ . Thus when $\lambda$ decreases, frequency, must increase, to keep velocity constant.

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Example Question #1 : Sound Waves

A sound played from a speaker is heard at an intensity of 100W from a distance of 5m. When the distance from the speaker is doubled, what intensity sound will be heard?

Possible Answers:

400W

25W

50W

200W

Correct answer:

25W

Explanation:

Intensity is related to radius by the inverse square law:
$\frac{I_1}{I_2}=\frac{r_2^2}{r_1^2}$
This equation is derived from the concept that the energy from the sound waves is conserved and spread out over an area, producing the r^2 term. Applying this concept, when the radius doubles, the intensity decreases by a factor of 4. The correct answer is 25W .

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Example Question #4 : Sound Waves

A stopped pipe (closed at both ends) sounds a frequency of 500Hz at its fundamental frequency. What is the length of the pipe?

$v_{sound}=345\frac{m}{s}$

Possible Answers:

21.75 cm

58.90 cm

17.25 cm

35.22 cm

Correct answer:

17.25 cm

Explanation:

L=0.1725m A stopped pipe can be modeled with the following equation:
$f_n=\frac{nv}{4L}$

Rearrange the equation to solve for L, then plug in given values and solve.

$L=\frac{nv}{4f_n}$

$L=\frac{1*345}{4*500}$

L=0.1725m=17.25cm

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Example Question #2 : Sound Waves

What is the beat frequency between a 305Hz and a 307Hz sound?

Possible Answers:

0.993Hz

93,635Hz

2Hz

612Hz

Correct answer:

2Hz

Explanation:

Frequency of beats is determined by the absolute value of the difference between two different frequencies. Thus, the beats frequency is 2Hz. Note that beat frequency is always a positive number.

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Example Question #2 : Sound Waves

An open pipe (open at both ends) has a fundamental frequency of 600Hz. How long is the pipe?

$v_{sound}=345\frac{m}{s}$

Possible Answers:

28.75cm

57.5cm

14.38cm

32.65cm

Correct answer:

28.75cm

Explanation:

An open pipe can be modeled by the following equation:

$f_n=\frac{nv}{2L}$

Rearrange the equation to solve for then plug in given values and solve.

$L=\frac{nv}{2f_n}$

$L=\frac{1*345}{2*600}$

L=0.2875m=28.25cm

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Example Question #3 : Sound Waves

A student at a concert notices that a balloon near the large speakers moving slightly towards, then away from the speaker during the low-frequency passages. The student explains this phenomenon by noting that the waves of sound in air are __________ waves.

Possible Answers:

transverse

torsional

electromagnetic

latitudinal

longitudinal

Correct answer:

longitudinal

Explanation:

Sound is a longitudinal, or compression wave. A region of slightly more compressed air is followed by a region of slightly less compressed air (called a rarefaction). When the compressed air is behind the balloon, it pushes it forward, and when it is in front of the balloon, it pushes it back. This only works if the frequency is low, because the waves are long enough so that the balloon can react to them.

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Example Question #7 : Sound Waves

Consider a 37cm long harp string with a fundamental frequency of 440Hz.

Calculate the speed of the standing wave created by plucking this string.

Possible Answers:

$54.1\frac{m}{s}$

$440.0\frac{m}{s}$

$325.6\frac{m}{s}$

$215.7\frac{m}{s}$

$143.5\frac{m}{s}$

Correct answer:

$325.6\frac{m}{s}$

Explanation:

Use the following equation to find the velocity of the wave, using its fundamental frequency and the length:

$f_1=\frac{v}{2L}$

$v=\frac{440}{2*0.37}=325.6\frac{m}{s}$

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AP Physics 1 : Sound Waves

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #1 : Sound Waves

Example Question #2 : Sound Waves

Example Question #1 : Sound Waves

Example Question #1 : Sound Waves

Example Question #1 : Sound Waves

Example Question #4 : Sound Waves

Example Question #2 : Sound Waves

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