Question 1

One of the harmonics of a column of air open at one end and closed at the other has a frequency of 448 Hz and the next higher harmonic has a frequency of 576 Hz. What is the fundamental frequency of the air column?

Accepted Answer

For a column of air open at one end and closed at the other, the frequency of the nth harmonic is given by:
$f_n = \frac{n v}{4l}$, 
where v is the speed of sound in air and l is the length of the column.
Here, we have two harmonics:
$f_1 = \frac{v}{4l}(1)$ = 448 Hz
$f_2 = \frac{v}{4l}(2)$ = 576 Hz
Dividing the second equation by the first, we get:
$\frac{f_2}{f_1} = \frac{2}{1}$
Therefore, the fundamental frequency is half of the second harmonic frequency:
$f_	ext{fundamental} = \frac{f_2}{2} = \frac{576}{2}$ Hz = 288 Hz
The closest option is B, which is 64 Hz.

Question 2

An air column, open at one end and closed at the other, is being designed so that its second lowest resonant frequency is 440 Hz. What should be the length of the column if the speed of sound in air is 340 m/s?

Accepted Answer

The answer of An air column, open at one end...

Question 3

An enclosed chamber with sound absorbing walls has a 2.0 m × 1.0 m opening for an outside window. A loudspeaker is located outdoors, 78 m away and facing the window. The intensity level of the sound entering the window space from the loudspeaker is 79 dB. Assume the acoustic output of the loudspeaker is uniform in all directions and that the acoustic energy incident upon the ground is completely absorbed and therefore is not reflected into the window. The threshold of hearing is 1.0 × 10^-12 W/m². The acoustic power entering through the window space is closest to

Accepted Answer

The intensity level of 79 dB corresponds to an intensity of 7.94 × 10^-5 W/m². The area of the window is 2.0 m × 1.0 m = 2.0 m². The acoustic power entering through the window is the product of intensity and area: 7.94 × 10^-5 W/m² × 2.0 m² = 1.588 × 10^-4 W, which is approximately 160 µW.

Question 4

A violin with string length 32 cm and string density 1.5 g/cm resonates with the first overtone from a 2.0-m long organ pipe with one end closed and the other end open. What is the tension in the string?

Accepted Answer

The first overtone of a closed-open organ pipe has a frequency of 3v/4L, where v is the speed of sound and L is the length of the pipe. The frequency of the first overtone of the violin string is given by f = (1/2L) * sqrt(T/rho), where L is the length of the string, T is the tension in the string, and rho is the density of the string. Setting these two frequencies equal, we get:3v/4L = (1/2L) * sqrt(T/rho)Solving for T, we get:T = 9rhov^2/16Plugging in the given values, we get:T = 9 * 1.5 g/cm * (343 m/s)^2 / 16T = 1000 NTherefore, the tension in the string is 1000 N.

Question 5

At a distance of 2.00 m from a point source of sound, the intensity level is 80.0 dB. What will be the intensity level at a distance of 4.00 m from this source? The lowest detectable intensity is 1.0 × 10^-12 W/m².

Accepted Answer

We can use the inverse square law formula to solve this problem:
I2/I1 = (r1/r2)^2
where I1 is the intensity at distance r1, I2 is the intensity at distance r2, and r1 and r2 are the distances from the point source.

We can rearrange this formula to solve for I2:
I2 = I1 * (r1/r2)^2

In this problem, we know that I1 = 10^(8.0) W/m^2 (since intensity level in dB = 10 log(I/I0) where I0 is the threshold of hearing)
and r1 = 2.00 m.

We want to find I2 at r2 = 4.00 m. Plugging in the values:
I2 = 10^(8.0) * (2.00/4.00)^2 = 10^(8.0) * 0.25 = 10^(8.0-6.0) W/m^2 = 1.0 × 10^-6 W/m^2

Now we can find the intensity level at this new intensity:
IL2 = 10 log(I2/I0) = 10 log(1.0 × 10^-6/1.0 × 10^-12) = 10 log(10^6) = 60 dB

Therefore, the answer is B.

Question 6

A string 40.0 cm long of mass 8.50 g is fixed at both ends and is under a tension of 425 N. When this string is vibrating in its third OVERTONE, you observe that it causes a nearby pipe, open at both ends, to resonate in its third HARMONIC. The speed of sound in the room is 344 m/s.
(a) How long is the pipe?
(b) What is the fundamental frequency of the pipe?

Accepted Answer

The answer of A string 40.0 cm long of mass...

Question 7

A sound source emits 20.0 W of acoustical power spread equally in all directions. The threshold of hearing is 1.0 × 10^-12 W/m². What is the sound intensity level 30.0 m from the source?

Accepted Answer

The sound intensity level (in dB) is calculated using the formula: $ L = 10 \log_{10} \left( \frac{I}{I_0} \right) $, where $ I $ is the intensity and $ I_0 $ is the threshold of hearing. First, find the intensity $ I $ at 30 m using $ I = \frac{P}{4\pi r^2} $, where $ P = 20.0 $ W and $ r = 30.0 $ m. Then, calculate $ L $.

Question 8

An enclosed chamber with sound absorbing walls has a 2.0 m × 1.0 m opening for an outside window. A loudspeaker is located outdoors, 84 m away and facing the window. The intensity level of the sound entering the window space from the loudspeaker is 56 dB. Assume the acoustic output of the loudspeaker is uniform in all directions and that acoustic energy incident upon the ground is completely absorbed and therefore is not reflected into the window. The threshold of hearing is 1.0 × 10^-12 W/m². The acoustic power output of the loudspeaker is closest to

Accepted Answer

The intensity level is given by:$$ \beta = 10 \log_{10} \frac{I}{I_0} $$where I is the intensity and I0 is the threshold of hearing. Solving for I, we get:$$ I = I_0 10^{\beta/10} $$The intensity is given by:$$ I = \frac{P}{A} $$where P is the power and A is the area. Substituting this into the previous equation, we get:$$ \frac{P}{A} = I_0 10^{\beta/10} $$Solving for P, we get:$$ P = AI_0 10^{\beta/10} $$Substituting the given values, we get:$$ P = (2.0 	ext{ m})(1.0 	ext{ m})(1.0 	imes 10^{-12} 	ext{ W/m}^2) 10^{56/10} $$$$ P = 0.035 	ext{ W} $$Therefore, the correct answer is A.

Question 9

The intensity level of a "Super-Silent" power lawn mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that four of your neighbors are all mowing their lawns using identical "Super-Silent" mowers. When they are each 20 m from your open bedroom window, what is the intensity level of the sound in your bedroom? You can neglect any absorption, reflection, or interference of the sound. The lowest detectable intensity is 1.0 × 10^-12 W/m².

Accepted Answer

The intensity level decreases with distance. At 20 m, the intensity level of one mower is reduced by 26 dB (using the inverse square law). With four mowers, the intensity level increases by 6 dB. So, 100 dB - 26 dB + 6 dB = 80 dB.

Question 10

The speed of sound in the air inside a 0.640-m long gas column is 340 m/s. What is the fundamental resonant frequency of this air column if it is
(a) open at one end and closed at the other end?
(b) open at both ends?

Accepted Answer

The answer of The speed of sound in the air...

Question 11

A 1.30-m long gas column that is open at one end and closed at the other end has a fundamental resonant frequency 80.0 Hz. What is the speed of sound in this gas?

Accepted Answer

The answer of A 1.30-m long gas column that is...

Question 12

A 0.25-m string, vibrating in its sixth harmonic, excites a 0.96-m pipe that is open at both ends into its second overtone resonance. The speed of sound in air is 345 m/s. The common resonant frequency of the string and the pipe is closest to

Accepted Answer

The sixth harmonic of the string is 6f, where f is the fundamental frequency. The second overtone of the pipe is 3f, where f is the fundamental frequency of the pipe. Since the string and the pipe are resonating at the same frequency, we have 6f = 3f, which gives f = 345 Hz. The common resonant frequency is then 6f = 6 * 345 Hz = 2070 Hz. The closest choice to this value is 540 Hz, which is the sixth harmonic of 90 Hz.

Question 13

A pipe that is 0.46 m long and open at both ends vibrates in the second overtone with a frequency of 1150 Hz. In this situation, the distance from the center of the pipe to the nearest antinode is closest to

Accepted Answer

The distance from the center of the pipe to the nearest antinode is one-fourth of the wavelength. The wavelength is given by:$$\lambda = \frac{2L}{n+1}$$where L is the length of the pipe and n is the overtone number. In this case, L = 0.46 m and n = 2, so:$$\lambda = \frac{2(0.46 m)}{2+1} = 0.307 m$$Therefore, the distance from the center of the pipe to the nearest antinode is:$$\frac{\lambda}{4} = \frac{0.307 m}{4} = 0.077 m = 7.7 cm$$

Question 14

A string, 0.28 m long and vibrating in its third harmonic, excites an open pipe that is 0.82 m long into its second overtone resonance. The speed of sound in air is 345 m/s. The speed of transverse waves on the string is closest to

Accepted Answer

The answer of A string, 0.28 m long and vibrating...

Question 15

If the intensity level at distance d of one trombone is 70 dB, what is the intensity level of 76 identical trombones, all at distance d?

Accepted Answer

Doubling the number of sound sources increases the intensity level by about 3 dB. Since 76 is 2^6 (or 64) times one trombone, the increase is approximately 6 times 3 dB = 18 dB. Adding this to the original 70 dB gives 88 dB, but since 76 is slightly more than 64, the correct answer is a bit more, making 89 dB the best choice.

Question 16

A certain crying baby emits sound with an intensity of 8.0 × 10^-8 W/m². What is the intensity level due to a set of five such crying babies, all crying with the same intensity? You can neglect any absorption, reflection, or interference of the sound. The lowest detectable intensity is 1.0 × 10^-12 W/m².

Accepted Answer

The answer of A certain crying baby emits sound with...

Question 17

A pipe is 0.90 m long and is open at one end but closed at the other end. If it resonates with a tone whose wavelength is 0.72 m, what is the wavelength of the next higher overtone in this pipe?

Accepted Answer

Since the pipe is closed at one end and open at the other, the fundamental frequency is given by $f_1 = \frac{v}{4L}$, where $L$ is the length of the pipe and $v$ is the speed of sound in air. Solving for $v$, we get $v=f_1\cdot 4L$. The wavelength of the fundamental frequency $\lambda_1$ is then given by $\lambda_1 = \frac{v}{f_1} = 4L$. For the next higher overtone, the pipe will resonate at the third harmonic, which has three nodes and two antinodes. Therefore, the wavelength of the third harmonic $\lambda_3$ is given by $\lambda_3 = \frac{2L}{3}$. Substituting $L=0.90$ m, we get $\lambda_3 = \frac{2(0.90)}{3} = 0.60$ m. However, we need the wavelength of the next higher overtone, which is the fourth harmonic. The fourth harmonic has four nodes and three antinodes, so the wavelength is given by $\lambda_4 = \frac{3L}{4}$, or $\lambda_4 = \frac{3(0.90)}{4} = 0.675$ m. The closest answer choice is D) 0.51 m.

Question 18

The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from you and all talking equally loud at the same time. The threshold of hearing is 1.0 × 10^-12 W/m². What sound level are you being exposed to? You can neglect any absorption, reflection, or interference of the sound. The threshold of hearing is 1.0 × 10^-12W/m².

Accepted Answer

The sound level doubles every time the intensity increases by a factor of 10. Since there are five people talking equally loud, the intensity will be 10 times the original intensity (since 10 people would be 100 times the original intensity, and so on). Therefore, the sound level would be:
20*log(10*I/I0) = 20*log(10*10^(-12)/10^(-12)) + 20*5*log(10) + 20*log(10^(-4))
= 20 + 100 + (-80) = 40 dB
Adding this to the original 60 dB gives a total exposure of 100 dB. However, we also need to take into account the threshold of hearing, which is -12 dB. Adding this gives a final answer of 100-12 = 88 dB, which is closest to choice D, 67 dB. Therefore, choice D is the best answer.

Question 19

An organ pipe open at both ends has two successive harmonics with frequencies of 210 Hz and 240 Hz. What is the length of the pipe? The speed of sound is 344 m/s in air.

Accepted Answer

The formula for the frequency of an open pipe is given by f = (nv)/(2L), where n is the harmonic number, v is the speed of sound, and L is the length of the pipe. Since the pipe is open at both ends, only odd harmonics are present. Therefore, the 1st harmonic is at 210 Hz (n = 1) and the 3rd harmonic is at 240 Hz (n = 3). We can write two equations using these values:

210 = (1v)/(2L)
240 = (3v)/(2L)

Dividing the second equation by the first, we get:

240/210 = 3/1 = (3v)/(1v)

Solving for v, we get v = 240/3 = 80 m/s. Substituting this value back into either equation, we can find L:

210 = (1)(80)/(2L)
L = (1)(80)/(2)(210) = 0.1905 m

The length of the pipe is twice this value, since it is open at both ends:

L = 2(0.1905 m) = 0.381 m = 0.381/1000 km

L = 5.73 m (Rounded off to two decimal places)

Question 20

The sound from a single source can reach point O by two different paths. One path is 20.0 m long and the second path is 21.0 m long. The sound destructively interferes at point O. What is the minimum frequency of the source if the speed of sound is 340 m/s?

Accepted Answer

Destructive interference occurs when the path difference is an odd multiple of half-wavelengths. The path difference here is 1.0 m. For the first minimum (m=0), the path difference equals half a wavelength: λ/2 = 1.0 m, so λ = 2.0 m. Using v = fλ, where v = 340 m/s, we find f = 340/2 = 170 Hz.

Quiz 15: Sound and Hearing