AP Physics 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?

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 term. Applying this concept, when the radius doubles, the intensity decreases by a factor of 4. The correct answer is .

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

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.

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

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

$325.6\frac{m}{s}$

$215.7\frac{m}{s}$

$54.1\frac{m}{s}$

$143.5\frac{m}{s}$

$440.0\frac{m}{s}$

Explanation

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

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

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

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.

The shorter length strings produce higher frequencies

The shorter length strings produce lower frequencies

The shorter length creates a higher speed of sound

The shorter length creates a lower speed of sound

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.

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?

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.

A guitar player hits a wrong note. The note he's supposed to hit is $1045\textup{ Hz}$ , but he's sharp and the note turns out to be $1058\textup{ Hz}$ . Starting from rest, how fast (and in which direction) do you have to run until the note sounds correct to you?

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

$4.21\ \frac{\textup{m}}{\textup{s}}$ away from the noise.

$4.21\ \frac{\textup{m}}{\textup{s}}$ towards the noise.

$37\ \frac{\textup{m}}{\textup{s}}$ towards the noise.

$37\ \frac{\textup{m}}{\textup{s}}$ away from the noise.

$16\ \frac{\textup{m}}{\textup{s}}$ away from the noise.

Explanation

In order to find the velocity needed, we use the Doppler equation:

$f_{observed}=\left(\frac{v\pm v_o}{v\mp v_s}\right)f_{source}$

Where is the speed of sound in air, is the velocity of the observer, and is the velocity of the source, which is zero in this case since the guitar player is not moving. Since the guitarist is sharp, your adjustment must decrease the perceived frequency by $13\textup{ Hz}$ , which means we use the negative sign in the numerator of the Doppler equation. Plug in the given values into the equation and solve for .

$1045Hz=\left(\frac{343\frac{m}{s}- v_o}{343\frac{m}{s}}\right)1058Hz$

$-4459\frac{m}{s^2}=-1058Hz*v_o$

$v_o=4.21\frac{m}{s}$

Because we need the note to be the final note to be lower in frequency than the original frequency, we need to run away from it.

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.

longitudinal

transverse

torsional

electromagnetic

latitudinal

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.

The speed of sounds is the fastest in which of the following media?

Air

Water

Glass

Vacuum

The speed of sound is constant in all media.

Explanation

The speed of sound is fastest in the least compressible media of the lowest density. Sound does not propagate in a vacuum. Air and water are compressible media, so sound does not travel as fast in these as it does in glass, an incompressible medium. In general, the speed of sound is greatest in solids, and within each phase, faster as density decreases.

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

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

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.

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?

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

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 is the initial frequency of the sound.