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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #1 : Electricity And 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:

57.5cm

28.75cm

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 #1 : 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:

electromagnetic

longitudinal

latitudinal

torsional

transverse

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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Example Question #11 : Sound Waves

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

Suppose the string is pressed down in such a way that only a 10cm length of string vibrates. What is the speed of the wave produced when the string is plucked in terms of $v_{i}$ the speed of the wave when all 37cm of the string vibrate?

Possible Answers:

$v_{i}$

$v_{i}-\frac{10}{440}$

$\frac{v_{i}}{10}$

$\frac{v_i}{17}$

$\frac{17v_{i}}{37}$

Correct answer:

$v_{i}$

Explanation:

The speed of a wave is a property of the medium, it is not affected by the length of the string. The frequency may change, but the speed remains constant. There is no change in the speed.

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Example Question #1161 : Ap Physics 1

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

If only half of the string is allowed to vibrate, what frequency will be heard?

Possible Answers:

126Hz

440Hz

220Hz

110Hz

880Hz

Correct answer:

880Hz

Explanation:

Since the speed of the wave does not change based on the length of the string and we know it has a fundamental frequency of 440Hz, a string of half the length will vibrate at twice the frequency, 880Hz. This makes sense as it will sound higher in pitch. You can try this with a rubber band on a shoebox. Plucking it while placing your finger halfway along the band will result in a higher pitched sound.

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Example Question #1162 : Ap Physics 1

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

What is the wavelength of the second harmonic of the string?

Possible Answers:

220cm

37cm

44cm

18.5cm

54cm

Correct answer:

37cm

Explanation:

The wavelength of the second harmonic of a standing wave on a string is just the length of the string. For the second harmonic, an entire cycle occurs on the length of the string. Therefore, the wavelength of the second harmonic for this string is 37cm or 0.37m. The wavelength for the first harmonic, or fundamental, is twice the length of the string, as this is when one half a cycle occurs over the length of the string.

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Example Question #14 : Electricity And Waves

If the first harmonic of a string has frequency , what is the frequency of the $n^{th}$ harmonic of that string in terms of ?

Possible Answers:

n+f

$\frac{f}{n}$

f^n

Correct answer:

Explanation:

Harmonics describe the relationship between wavelengths and frequency on a string. A given string will have a wave speed associated with it. Given this wave speed, the harmonics are the frequencies at which half-wavelengths occur along the length of the string. For instance, the first harmonic is the frequency at which a half-wavelength occurs over the string. The second harmonic completes one wave over the string. The string is 1.5 times the wavelength of the third harmonic and so on. Therefore, the $n^{th}$ harmonic has a frequency of .

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Example Question #11 : Sound Waves

Given that 20kJ of energy are hitting a window pane over a period of 5s with dimensions 2.5m by 4m, what is the sound level in decibels?

$I_o=10^{-12}\frac{W}{m^2}$

Possible Answers:

140log(4)dB

$10log(4*10^{14})dB$

$10log(2*10^{15})dB$

14log(4)dB

Correct answer:

$10log(4*10^{14})dB$

Explanation:

The formula for intensity, is:

$I=\frac{P}{A}$

Where is power in watts and is the surface area. The surface area in our case is:

A=(2.5*4)m^2=10m^2

The power can be given as $\frac{J}{s}$ or in our case:

$P= \frac{2*10^4J}{5s}=4*10^3 W$

Solve for intensity.

$I=\frac{P}{A}= \frac{4*10^3W}{10m^2}= 400 \frac{W}{m^2}$

The formula for sound level is:

$B=10log\left(\frac{I}{I_o}\right)$ , where is the intensity and I_o is the threshold of hearing, which is $10^{-12}\frac{W}{m^2}$

$B=10log\left(\frac{4*10^2}{10^{-12}}\right)=10log(4*10^{14}) dB$

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Example Question #14 : Sound Waves

At a distance of from a fan exerting 50W of mechanical energy, estimate the sound level if the threshold of hearing is $10^{-12}\frac{W}{m^2}$

Possible Answers:

$.884\:dB$

$110\:dB$

$144\:dB$

$114\:dB$

Correct answer:

$114\:dB$

Explanation:

First we need to solve for intensity , given by:

$I=\frac{P}{4\pi r^2}$ , where is power, is the distance from the source of the sound.

In our case, P=50W and r=4 , therefore

$I=\frac{50W}{4\pi (4m^2)}=\frac{25}{32\pi } \frac{W}{m^2}\approx.25\frac{W}{m^2}=2.5\times10^{-1}\frac{W}{m^2}$

To solve for sound level , we do

$B=10\log(\frac{I}{I_o})$ , where is the intensity and I_o is the threshold of hearing.

In this problem,

$I=2.5\times10^{-1}\frac{W}{m^2}$ , and $I_o=10^{-12}\frac{W}{m^2}$

$B=10\log(\frac{2.5\times10^{-1}}{10^{-12}}})= 114\:dB$

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Example Question #17 : Electricity And Waves

By what factor will the sound level in decibels change if the intensity is increased by a factor of ?

Possible Answers:

Sound level will change by a factor of $log\left(\frac{1}{2}\right)$

Sound level will change by a factor of 100

Sound level will change by a factor of log(2)

Sound level will change by a factor of

Correct answer:

Sound level will change by a factor of log(2)

Explanation:

Recall that the formula for sound level given in decibels is given by:

$B=10log\left(\frac{I}{I_0}\right)$ , where is the intensity and I_0 is the threshold of hearing.

Sound level is proportional to intensity by:

$B\propto log(I)$

If the intensity is increased by a factor of , sound level would increase by a factor of log(2)

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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #1 : Electricity And Waves

Example Question #1 : Sound Waves

Example Question #7 : Sound Waves

Example Question #11 : Sound Waves

Example Question #1161 : Ap Physics 1

Example Question #1162 : Ap Physics 1

Example Question #14 : Electricity And Waves

Example Question #11 : Sound Waves

Example Question #14 : Sound Waves

Example Question #17 : Electricity And Waves

All AP Physics 1 Resources

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