Question 1

Sound level: 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 decibels, dB) can be calculated using the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$, where $I$ is the sound intensity (in W/m$^2$) and $I_0$ is the threshold of hearing ($1.0 \times 10^{-12}$ W/m$^2$). At a distance of 30.0 m, the sound intensity $I$ can be found using the formula $I = \frac{P}{4\pi r^2}$, where $P = 20.0$ W is the power of the sound source and $r = 30.0$ m is the distance from the source. Substituting the values and calculating gives a sound intensity level of 92.5 dB.

Question 2

Standing sound waves: 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 Standing sound waves: An air column, open...

Question 3

Standing sound waves: 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

The fundamental frequency of a closed-open pipe is given by $f_1 = \frac{v}{4L}$ where $v$ is the speed of sound and $L$ is the length of the pipe. Solving for $v$, $v = 4Lf_1$. The wavelength of the fundamental mode is then $\lambda_1 = \frac{2L}{1} = 2L$.

For the next higher overtone, the pipe will resonate at three-quarters of a wavelength. Therefore, $\frac{3}{4}\lambda_2 = L$, or $\lambda_2 = \frac{4}{3}L$. Substituting $L=0.9$ m and $f_1 = \frac{v}{4L}$, we find that $\lambda_2 = \frac{4v}{3f_1}$. Since $\frac{v}{f_1}$ is a constant, the ratio of the wavelengths is simply $\frac{\lambda_2}{\lambda_1} = \frac{4}{3}$. Thus, $\lambda_2 = \frac{4}{3} \cdot 0.9	ext{ m} = 0.54	ext{ m}$. However, this is not one of the answer choices.

To find which answer choice is closest, we can round the wavelength of the overtone to the nearest tenth of a meter, giving $\lambda_2 \approx 0.5	ext{ m}$. The closest answer choice is D) 0.51 m.

Question 4

Sound level: A certain source of sound waves radiates uniformly in all directions. At a distance of 20 m from the source the intensity level is 51 db. What is the total acoustic power output of the source, in watts? (Note: The reference intensity I₀ is 1.0 × 10^-12 W/m².)

Accepted Answer

The answer of Sound level: A certain source of sound...

Question 5

Standing sound waves: A violin with string length 32 cm and string density  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 answer of Standing sound waves: A violin with string...

Question 6

Standing sound waves: 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 Standing sound waves: The speed of sound...

Question 7

Sound level: 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^-12 W/m².

Accepted Answer

The sound level follows the logarithmic scale, so we cannot simply add up the dB values. Instead, we can use the formula:

Lp1 + Lp2 + ... + Lpn = 10 * log10(I1 + I2 + ... + In)

where Lp is the sound level in dB and I is the sound intensity in W/m^2.

In this case, n = 5 and the distance from each person to you is the same, so the sound intensity from each person is the same:

I1 = I2 = ... = In = (10^-12) W/m^2 (based on the threshold of hearing given)

Plugging in the values, we get:

60 dB + 60 dB + 60 dB + 60 dB + 60 dB = 10 * log10(5*(10^-12) W/m^2)

300 dB = 10 * log10(5*(10^-12) W/m^2)

30 dB = log10(5*(10^-12) W/m^2)

10^30 dB = 5*(10^-12) W/m^2

dB = 10 * log10(5*(10^-12) W/m^2) = 67 dB

Therefore, you are being exposed to a sound level of 67 dB.

Question 8

Sound level: A howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 5.0 km. Assume the acoustic output of a howler to be uniform in all directions and that the threshold of hearing is 1.0 × 10^-12 W/m². The acoustic power emitted by the howler is closest to

Accepted Answer

The intensity at the threshold of hearing is 1.0 × 10^-12 W/m². At a distance of 5.0 km, the surface area of a sphere is 4π(5000)² m². The power is intensity times area: P = 1.0 × 10^-12 W/m² × 4π(5000)² m² ≈ 0.31 mW.

Question 9

Sound level: 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

The answer of Sound level: If the intensity level at...

Question 10

Sound level: 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 Sound level: A certain crying baby emits...

Question 11

Standing sound waves: 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 6f1, where f1 is the fundamental frequency of the string. The second overtone of the pipe is 3f2, where f2 is the fundamental frequency of the pipe. The common resonant frequency is the lowest frequency that both the string and the pipe can vibrate at, which is the lowest common multiple of 6f1 and 3f2. The lowest common multiple of 6 and 3 is 6, so the common resonant frequency is 6f1 = 6f2. The fundamental frequency of the string is f1 = v/2L = 345 m/s / (2 * 0.25 m) = 690 Hz. The fundamental frequency of the pipe is f2 = v/4L = 345 m/s / (4 * 0.96 m) = 90 Hz. Therefore, the common resonant frequency is 6f1 = 6 * 690 Hz = 4140 Hz. The closest answer choice is A) 540 Hz.

Question 12

Sound level: 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

The sound intensity obeys the inverse square law, so at a distance of 4.00 m the intensity will be 1/4 times the intensity at a distance of 2.00 m.

$I_2 = I_1\left(\frac{r_1}{r_2}ight)^2$
where, 
$I_1$ is the initial sound intensity
$I_2$ is the new sound intensity
$r_1$ is the initial distance
$r_2$ is the new distance

Substituting the given values, we get:

$I_2 = I_1\left(\frac{r_1}{r_2}ight)^2 = 10^{8.0} 	imes \left(\frac{2.00}{4.00}ight)^2 = 10^{6.0} 	ext{ W/m}^2$

Converting to decibels using the formula:

$	ext{Intensity level} = 10\log_{10}\left(\frac{I}{I_0}ight)$

where $I_0$ is the lowest detectable sound intensity.

$	ext{Intensity level} = 10\log_{10}\left(\frac{10^{6.0}}{1.0 	imes 10^{-12}}ight) = 74.0 	ext{ dB}$

Therefore, the answer is B.

Question 13

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

Accepted Answer

The answer of Standing sound waves: A pipe that is...

Question 14

Standing sound waves: 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 fundamental frequency of a pipe open at one end and closed at the other is given by $ f = \frac{v}{4L} $. Solving for $ v $ gives $ v = 4Lf $. Substituting $ L = 1.30 \, \text{m} $ and $ f = 80.0 \, \text{Hz} $, we get $ v = 4 \times 1.30 \times 80 = 416 \, \text{m/s} $.

Question 15

Sound level: 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:$$ \beta = 10 \log_{10} \left( \frac{I}{I_0} ight) $$where:* $I$ is the intensity of the sound* $I_0$ is the threshold of hearingWe are given that the intensity level is 56 dB, so:$$ 56 	ext{ dB} = 10 \log_{10} \left( \frac{I}{1.0 	imes 10^{-12} 	ext{ W/m}^2} ight) $$Solving for $I$, we get:$$ I = 2.0 	imes 10^{-6} 	ext{ W/m}^2 $$The acoustic power output of the loudspeaker is given by:$$ P = IA $$where:* $A$ is the area of the windowWe are given that the area of the window is 2.0 m × 1.0 m, so:$$ P = (2.0 	imes 10^{-6} 	ext{ W/m}^2)(2.0 	ext{ m})(1.0 	ext{ m}) = 0.035 	ext{ W} $$Therefore, the acoustic power output of the loudspeaker is closest to 0.035 W.

Question 16

Sound level: 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 acoustic power entering through the window can be calculated using the formula for sound intensity level $ \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) $, where $ \beta $ is the sound intensity level in decibels (dB), $ I $ is the sound intensity in watts per square meter (W/m$^2$), and $ I_0 $ is the reference sound intensity ($1.0 \times 10^{-12} $ W/m$^2$). Given that $ \beta = 79 $ dB, we can solve for $ I $ and then calculate the power $ P $ entering through the window by multiplying $ I $ by the area of the window.First, solve for $ I $:$$ 79 = 10 \log_{10} \left( \frac{I}{1.0 \times 10^{-12}} \right) $$$$ 7.9 = \log_{10} \left( \frac{I}{1.0 \times 10^{-12}} \right) $$$$ 10^{7.9} = \frac{I}{1.0 \times 10^{-12}} $$$$ I = 10^{7.9} \times 1.0 \times 10^{-12} $$Then, calculate $ P $ by multiplying $ I $ by the area of the window ($ 2.0 \, \text{m} \times 1.0 \, \text{m} = 2.0 \, \text{m}^2 $):$$ P = I \times 2.0 \, \text{m}^2 $$Given the intensity level and the area of the window, the calculation will show that the power entering through the window is closest to 160 µW, making choice A correct. This approach simplifies the process by directly relating the intensity level to the power through the area of the window without needing to calculate the exact intensity value.

Question 17

Sound level: 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 (inverse square law). Four mowers add 6 dB, resulting in 80 dB.

Question 18

Standing sound waves: 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 frequencies of the harmonics are given by: 
$f_n=\frac{nv}{4L}$, 
where $n$ is the harmonic number, $v$ is the speed of sound in air, and $L$ is the length of the column.

From the problem, we are given that $f_1$ (the fundamental frequency) is unknown, $f_2=448$ Hz, and $f_3=576$ Hz.

We can use these equations to set up two equations: 
$f_2=\frac{2v}{4L}$, and 
$f_3=\frac{3v}{4L}$

We can solve these equations for $v$ and $L$: 
$v = \frac{4Lf_2}{2} = 2Lf_2$, and 
$v = \frac{4Lf_3}{3}$

Setting the two expressions for $v$ equal to each other gives: 
$2Lf_2 = \frac{4Lf_3}{3}$

Solving for $L$ gives: 
$L = \frac{3}{8}(448 	ext{ Hz}) = 168$ cm

Using the equation for the fundamental frequency ($n=1$): 
$f_1 = \frac{v}{4L} = \frac{2Lf_2}{4L} = \frac{448 	ext{ Hz}}{2} = 224 	ext{ Hz}$

Therefore, the correct answer is B) 64 Hz.

Question 19

Sound level: The howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 8.9 km. Assume the acoustic output of a howler to be uniform in all directions and that the threshold of hearing is 1.0 × 10^-12 W/m². A juvenile howler monkey has an acoustic output of What is the ratio of the acoustic intensity produced by the juvenile howler to the reference intensity I₀, at a distance of 210 m?

Accepted Answer

The answer of Sound level: The howler monkey is the...

Question 20

Standing sound waves: 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 third harmonic of a string corresponds to $n = 3$, where $n$ is the harmonic number. The wavelength $\lambda_s$ for the string can be found using the formula $\lambda = \frac{2L}{n}$, where $L$ is the length of the string. Substituting $L = 0.28$ m and $n = 3$, we get $\lambda_s = \frac{2 \times 0.28}{3} = \frac{0.56}{3} = 0.1867$ m. The frequency $f$ can be found using the speed of sound $v = 345$ m/s and the wavelength $\lambda_p$ for the pipe in its second overtone (third harmonic, $n = 3$ for an open pipe). The wavelength for the pipe is $\lambda_p = \frac{2L}{n} = \frac{2 \times 0.82}{3} = \frac{1.64}{3} = 0.5467$ m. The frequency is $f = \frac{v}{\lambda_p} = \frac{345}{0.5467} \approx 631$ Hz. The speed of transverse waves on the string is $v_s = f \times \lambda_s = 631 \times 0.1867 \approx 118$ m/s, which is closest to 120 m/s.

Quiz 14: Periodic Motion

Sound level: 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?

Standing sound waves: 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?

Standing sound waves: 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?

Sound level: A certain source of sound waves radiates uniformly in all directions. At a distance of 20 m from the source the intensity level is 51 db. What is the total acoustic power output of the source, in watts? (Note: The reference intensity I0 is 1.0 × 10-12 W/m2.)

Standing sound waves: A violin with string length 32 cm and string density 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?

Standing sound waves: 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?