Question 1

A string that is 0.15 m long and fixed at both ends is vibrating in its n = 5 harmonic.The sound from this string excites a pipe that is 
Long and open at both ends into its second overtone resonance.What is the distance between a node and an adjacent antinode,in the string? The speed of sound in air is 345 m/s.

Accepted Answer

The answer of A string that is 0.15 m long...

Question 2

The wavelengths of the sounds produced by two horns are 6.0 m and 7.0 m,respectively.What beat frequency is heard when the horns are sounded on a day when the speed of sound is 340 m/s?

Accepted Answer

The answer of The wavelengths of the sounds produced by...

Question 3

A string that is 0.26 m long is vibrating in its n = 6 harmonic.The sound from this string excites a pipe that is 
Long and open at both ends into its second overtone resonance.What is the common resonant frequency of the string and the pipe? The speed of sound in air is 345 m/s.

Accepted Answer

The answer of A string that is 0.26 m long...

Question 4

One of the harmonics of a column of air in a tube that is 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.How long is the tube? The speed of sound in air is 343 m/s.

Accepted Answer

The harmonics of a tube open at one end and closed at the other are odd multiples of the fundamental frequency. If one harmonic is 448 Hz and the next higher harmonic is 576 Hz, the difference between them represents two harmonic steps (since even harmonics are absent in such a tube). This means the fundamental frequency (first harmonic) can be found by dividing the difference by 2. The difference is 576 Hz - 448 Hz = 128 Hz. Dividing by 2 gives us 64 Hz for the fundamental frequency. The wavelength of the fundamental frequency ($\lambda_1$) is four times the length of the tube ($L$), because a quarter wavelength fits in the tube. So, $\lambda_1 = 4L$. The speed of sound ($v$) is the product of the frequency ($f$) and the wavelength ($\lambda$), so $v = f\lambda$. Substituting the values for the fundamental frequency and rearranging to solve for $L$, we get $L = \frac{v}{4f} = \frac{343 \, \text{m/s}}{4 \times 64 \, \text{Hz}} = 1.34 \, \text{m}$.

Question 5

Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.00 m.A person is standing 12.0 m away from the wall,equidistant from the loudspeakers.When the person moves 1.00 m parallel to the wall,she experiences destructive interference for the first time.What is the frequency of the sound? The speed of sound in air is 343 m/s.

Accepted Answer

The condition for destructive interference is that the path difference between the two waves should be an odd multiple of half wavelengths, i.e., $\Delta L = (2n + 1)\frac{\lambda}{2}$, where $n$ is an integer starting from 0 (for the first occurrence of destructive interference, $n = 0$). The path difference $\Delta L$ is created when the person moves parallel to the wall. Given the geometry, we can use the Pythagorean theorem to find the distances from the person to each speaker before and after the move, and then find the difference in these distances to get $\Delta L$.Before moving, the distance from the person to each speaker is the same. After moving 1.00 m parallel to the wall, the distances to the speakers become $d_1$ and $d_2$, with one being longer and the other shorter than the original distance. The original distance to each speaker is $\sqrt{12^2 + (5/2)^2} = \sqrt{144 + 6.25} = \sqrt{150.25} \approx 12.26$ m. After moving, the distances are $\sqrt{12^2 + (3/2)^2} = \sqrt{144 + 2.25} = \sqrt{146.25} \approx 12.09$ m and $\sqrt{12^2 + (7/2)^2} = \sqrt{144 + 12.25} = \sqrt{156.25} \approx 12.50$ m. The path difference $\Delta L$ is thus approximately $12.50 - 12.09 = 0.41$ m.Since this is the first occurrence of destructive interference, we have $\Delta L = \frac{\lambda}{2}$. Therefore, $\lambda = 2\Delta L = 2 \times 0.41 = 0.82$ m. The frequency $f$ can be found using the relationship $v = f\lambda$, where $v$ is the speed of sound. Solving for $f$, we get $f = \frac{v}{\lambda} = \frac{343}{0.82} \approx 418.29$ Hz, which is closest to the given option C) 422 Hz.

Question 6

A 0.13-m string,fixed at both ends and vibrating in its n = 4 harmonic,excites a pipe that is 0.88 m.  Long and open at both ends,into its second overtone resonance.What is the speed of transverse waves on the string? The speed of sound in air is 345 m/s.

Accepted Answer

- The fundamental frequency of a string fixed at both ends is given by f = (n/2L) * v, where n is the harmonic number, L is the length of the string, and v is the speed of transverse waves on the string.
- For the n = 4 harmonic, we have f = (4/2*0.13) * v = 15.38 v.
- The second overtone resonance of an open pipe is given by f = 3v/4L, where L is the length of the pipe and v is the speed of sound in air.
- For the second overtone resonance, we have f = 3v/4*0.88 = 1.03v.
- Equating these two frequencies gives 15.38 v = 1.03v, or v = 38 m/s.
- Therefore, the best choice is A) 38 m/s.

Question 7

Two in-phase loudspeakers are placed along a wall and are separated by a distance of 4.00 m.They emit sound with a frequency of 514 Hz.A person is standing away from the wall,in front of one of the loudspeakers.What is the closest nonzero distance from the wall the person can stand and hear constructive interference? The speed of sound in air is 343 m/s.

Accepted Answer

In order to get constructive interference, the path difference between the two waves should be an integer multiple of the wavelength. Let the distance of the person from the wall be x. Then, the path difference between the two waves reaching the person is given by 2xsinθ, where θ is the angle between the line joining the two speakers and the line joining the person and the wall. In this case, θ can be approximated to be the angle of incidence, which is given by θ = tan^{-1}(x/4). For constructive interference, we need 2xsinθ = nλ, where n is an integer. Rearranging, we get x = (nλ)/(2sinθ). Substituting the given values, we get x = 0.73 m for n = 1 (the closest nonzero distance). Therefore, the answer is B.

Question 8

Two taut strings of identical mass and length are stretched with their ends fixed,but the tension in one string is 1.10 times greater than in the other.Waves on the string with the lower tension propagate at 35.2 m/s.The fundamental frequency of that string is 258 Hz.What is the beat frequency when each string is vibrating at its fundamental frequency?

Accepted Answer

The answer of Two taut strings of identical mass and...

Question 9

Consider two pipes of the same length: one pipe is open at both ends and the other pipe is closed on one end but open at the other end.If the fundamental frequency of the totally open pipe is 300 Hz,what is the fundamental frequency of the other pipe?

Accepted Answer

The fundamental frequency of a pipe open at both ends is given by $f = \frac{v}{2L}$, where $v$ is the speed of sound and $L$ is the length of the pipe. For a pipe closed at one end, the fundamental frequency is given by $f = \frac{v}{4L}$, which is half the frequency of the open pipe because the closed pipe supports only odd harmonics. Therefore, if the open pipe has a fundamental frequency of 300 Hz, the closed pipe's fundamental frequency will be half of that, which is 150 Hz.

Question 10

A violin with string length 32 cm and string density 
Resonates in its fundamental with the first overtone of a 2.0-m organ pipe with one end closed and one end open.What is the tension in the string if the speed of sound in air is 344 m/s?

Accepted Answer

The answer of A violin with string length 32 cm...

Question 11

One of the harmonics of a column of air in a tube that is 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 in this tube?

Accepted Answer

The fundamental frequency (first harmonic) for a tube open at one end and closed at the other is given by $f = \frac{v}{4L}$, where $v$ is the speed of sound and $L$ is the length of the tube. The harmonics in such a tube occur at odd multiples of the fundamental frequency (i.e., $f, 3f, 5f, \ldots$). Given that one harmonic is 448 Hz and the next higher harmonic is 576 Hz, the difference between them represents two steps in odd harmonics (since even harmonics are absent in this setup). Therefore, the difference (128 Hz) is twice the fundamental frequency, making the fundamental frequency 64 Hz.

Question 12

One of the harmonics of a column of air in a tube that is open at both ends 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 in this tube?

Accepted Answer

The answer of One of the harmonics of a column...

Question 13

Two in-phase loudspeakers are 3.0 m apart.They emit sound with a frequency of 490 Hz.A microphone is placed half-way between the speakers and then moved along the line joining the two speakers until the first point of constructive interference is found.At what distance from that midpoint is that first point? The speed of sound in air is 343 m/s.

Accepted Answer

The answer of Two in-phase loudspeakers are 3.0 m apart.They...

Question 14

A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with a frequency of 39.6 kHz.A car is approaching him at a speed of 35.0 m/s.The wave is reflected by the car and interferes with the emitted sound producing beats.What is the frequency of the beats when the speed of sound in air is 343 m/s?

Accepted Answer

The beat frequency is calculated using the Doppler effect. The frequency of the reflected wave is higher than the emitted frequency due to the car's approach. The beat frequency is the difference between the frequency of the reflected wave and the emitted frequency. Calculating this gives a beat frequency of approximately 9000 Hz.

Question 15

Two loudspeakers placed 6.00 m apart are driven in phase by an audio oscillator having a frequency range from 1595 Hz to 2158 Hz.A point P is located 4.70 m from one loudspeaker and 3.60 m from the other speaker.The speed of sound in the room is 344 m/s.At what frequency (or frequencies)of the oscillator does the sound reaching point P interfere destructively? (There could be more than one correct choice. )

Accepted Answer

The path length difference between the two loudspeakers to point P is:$$6.00	ext{ m}-4.70	ext{ m}-3.60	ext{ m}= -2.30	ext{ m}$$For destructive interference, the path length difference must be an odd multiple of half a wavelength:$$\frac{1}{2}\lambda=\frac{c}{2f}$$$$2.30	ext{ m}=(2n+1)\frac{344	ext{ m/s}}{2f}$$$$f=\frac{344	ext{ m/s}}{4.60	ext{ m}}(2n+1)$$$$f=(74.8	ext{ Hz})(2n+1)$$The lowest frequency that produces destructive interference is:$$f=(74.8	ext{ Hz})(1)=74.8	ext{ Hz}$$This frequency is not within the given range, so we can ignore it.The next frequency that produces destructive interference is:$$f=(74.8	ext{ Hz})(3)=224.4	ext{ Hz}$$This frequency is also not within the given range, so we can ignore it.The next frequency that produces destructive interference is:$$f=(74.8	ext{ Hz})(5)=374.0	ext{ Hz}$$This frequency is within the given range, so it is a correct answer.The next frequency that produces destructive interference is:$$f=(74.8	ext{ Hz})(7)=523.6	ext{ Hz}$$This frequency is also within the given range, so it is a correct answer.The next frequency that produces destructive interference is:$$f=(74.8	ext{ Hz})(9)=673.2	ext{ Hz}$$This frequency is also within the given range, so it is a correct answer.The next frequency that produces destructive interference is:$$f=(74.8	ext{ Hz})(11)=822.8	ext{ Hz}$$This frequency is also within the given range, so it is a correct answer.The next frequency that produces destructive interference is:$$f=(74.8	ext{ Hz})(13)=972.4	ext{ Hz}$$This frequency is also within the given range, so it is a correct answer.The next frequency that produces destructive interference is:$$f=(74.8	ext{ Hz})(15)=1122.0	ext{ Hz}$$This frequency is also within the given range, so it is a correct answer.The next frequency that produces destructive interference is:$$f=(74.8	ext{ Hz})(17)=1271.6	ext{ Hz}$$This frequency is also within the given range, so it is a correct answer.The next frequency that produces destructive interference is:$$f=(74.8	ext{ Hz})(19)=1421.2	ext{ Hz}$$This frequency is also within the given range, so it is a correct answer.The next frequency that produces destructive interference is:$$f=(74.8	ext{ Hz})(21)=1570.8	ext{ Hz}$$This frequency is also within the given range, so it is a correct answer.The next frequency that produces destructive interference is:$$f=(74.8	ext{ Hz})(23)=1720.4	ext{ Hz}$$This frequency is also within the given range, so it is a correct answer.The next frequency that produces destructive interference is:$$f=(74.8	ext{ Hz})(25)=1870.0	ext{ Hz}$$This frequency is also within the given range, so it is a correct answer.The next frequency that produces destructive interference is:$$f=(74.8	ext{ Hz})(27)=2019.6	ext{ Hz}$$This frequency is also within the given range, so it is a correct answer.The next frequency that produces destructive interference is:$$f=(74.8	ext{ Hz})(29)=2169.2	ext{ Hz}$$This frequency is also within the given range, so it is a correct answer.Therefore, the correct answers are:$$f=374.0	ext{ Hz, }523.6	ext{ Hz, }673.2	ext{ Hz, }822.8	ext{ Hz, }972.4	ext{ Hz, }1122.0	ext{ Hz, }1271.6	ext{ Hz, }1421.2	ext{ Hz, }1570.8	ext{ Hz, }1720.4	ext{ Hz, }1870.0	ext{ Hz, }2019.6	ext{ Hz, }2169.2	ext{ Hz}$$However, the given range is from 1595 Hz to 2158 Hz, so the only correct answers within the given range are:$$f=1720.4	ext{ Hz, }2019.6	ext{ Hz}$$

Question 16

A 2.0-g string that is 0.67 m long is under tension.The string vibrates at 700 Hz tone in its third harmonic,and this vibration causes a sound wave.What is the wavelength of the sound? The speed of sound in air is 344 m/s.

Accepted Answer

The wavelength of the sound can be found using the formula for the speed of sound, $v = f\lambda$, where $v$ is the speed of sound in air (344 m/s), $f$ is the frequency of the sound (700 Hz), and $\lambda$ is the wavelength we need to find. Rearranging the formula to solve for $\lambda$, we get $\lambda = \frac{v}{f}$. Plugging in the given values, $\lambda = \frac{344 \, \text{m/s}}{700 \, \text{Hz}} = 0.49 \, \text{m}$.

Question 17

Two loudspeakers placed 6.00 m apart are driven in phase by an audio oscillator having a frequency range from 1908 Hz to 2471 Hz.A point P is located 4.70 m from one loudspeaker and 3.60 m from the other speaker.At what frequency of the oscillator does the sound reaching point P interfere constructively? The speed of sound is 344 m/s.

Accepted Answer

For constructive interference, the path difference should be an integer multiple of the wavelength. The path difference is 4.70 m - 3.60 m = 1.10 m. The wavelength λ is given by v/f, where v is the speed of sound (344 m/s). Solving for f using the condition for constructive interference (nλ = 1.10 m), we find that 2190 Hz is a frequency that satisfies this condition.

Question 18

Two taut strings of identical mass per unit length are stretched with the same tension with their ends fixed,but one string is 0.330 cm longer than the other.Waves on these strings propagate at 34.0 m/s.The fundamental frequency of the shorter string is 258 Hz.What is the beat frequency when each string is vibrating at its fundamental frequency?

Accepted Answer

The beat frequency is given by the difference in frequencies of the two strings. The frequency of a string is given by:$$f = \frac{v}{2L}\sqrt{\frac{T}{\mu}}$$where v is the wave speed, L is the length of the string, T is the tension, and $\mu$ is the mass per unit length.Since the two strings have the same mass per unit length and tension, the only difference in their frequencies is due to the difference in their lengths. The shorter string has a frequency of 258 Hz, so the longer string has a frequency of:$$f_2 = f_1\frac{L_1}{L_2} = 258 	ext{ Hz}\frac{L_1}{L_1 + 0.0033 	ext{ m}} = 258 	ext{ Hz}\frac{1}{1.0033} = 257.1 	ext{ Hz}$$The beat frequency is then:$$f_b = |f_1 - f_2| = |258 	ext{ Hz} - 257.1 	ext{ Hz}| = 0.9 	ext{ Hz}$$

Question 19

Two in-phase loudspeakers are placed along a wall and are separated by a distance of 4.00 m.They emit sound with a frequency of 514 Hz.A person is standing away from the wall,in front of one of the loudspeakers.What is the closest distance from the wall the person can stand and hear destructive interference? The speed of sound in air is 343 m/s.

Accepted Answer

Destructive interference occurs when the path difference is an odd multiple of half-wavelengths. The wavelength is calculated as speed/frequency = 343/514 ≈ 0.667 m. The path difference for the first instance of destructive interference is λ/2 = 0.667/2 ≈ 0.334 m, which corresponds to choice A.

Question 20

Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 8.00 m.A person is standing 12.0 m away from the wall,equidistant from the loudspeakers.When the person moves 3.00 m parallel to the wall,she experiences destructive interference for the second time.What is the frequency of the sound? The speed of sound in air is 343 m/s.

Accepted Answer

The distance between the first and second destructive interference is 6.00 m. This corresponds to 3 wavelengths, so the wavelength is 2.00 m. The frequency is then:f = v/λ = 343 m/s / 2.00 m = 171.5 Hz

Quiz 14: Waves and Sound

The wavelengths of the sounds produced by two horns are 6.0 m and 7.0 m,respectively.What beat frequency is heard when the horns are sounded on a day when the speed of sound is 340 m/s?

A string that is 0.26 m long is vibrating in its n = 6 harmonic.The sound from this string excites a pipe that is Long and open at both ends into its second overtone resonance.What is the common resonant frequency of the string and the pipe? The speed of sound in air is 345 m/s.

One of the harmonics of a column of air in a tube that is 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.How long is the tube? The speed of sound in air is 343 m/s.

A 0.13-m string,fixed at both ends and vibrating in its n = 4 harmonic,excites a pipe that is 0.88 m. Long and open at both ends,into its second overtone resonance.What is the speed of transverse waves on the string? The speed of sound in air is 345 m/s.

Consider two pipes of the same length: one pipe is open at both ends and the other pipe is closed on one end but open at the other end.If the fundamental frequency of the totally open pipe is 300 Hz,what is the fundamental frequency of the other pipe?

A violin with string length 32 cm and string density Resonates in its fundamental with the first overtone of a 2.0-m organ pipe with one end closed and one end open.What is the tension in the string if the speed of sound in air is 344 m/s?

One of the harmonics of a column of air in a tube that is 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 in this tube?

One of the harmonics of a column of air in a tube that is open at both ends 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 in this tube?

A 2.0-g string that is 0.67 m long is under tension.The string vibrates at 700 Hz tone in its third harmonic,and this vibration causes a sound wave.What is the wavelength of the sound? The speed of sound in air is 344 m/s.