Question 1

The fundamental frequency of a pipe that has one end closed is 256 Hz.When both ends of the same pipe are opened,the fundamental frequency is

Accepted Answer

The fundamental frequency of a pipe that has one end closed is given by:$$f_1 = \frac{v}{4L}$$where:* $f_1$ is the fundamental frequency* $v$ is the speed of sound in the pipe* $L$ is the length of the pipeWhen both ends of the pipe are opened, the fundamental frequency is given by:$$f_2 = \frac{v}{2L}$$Therefore, the fundamental frequency when both ends of the pipe are opened is twice the fundamental frequency when one end is closed. In this case, the fundamental frequency when one end is closed is 256 Hz, so the fundamental frequency when both ends are opened is 2 * 256 Hz = 512 Hz.The correct answer is B: 128 Hz.

Question 2

The third harmonic of a tube closed at one end is 735 Hz.If the speed of sound in air is 335 m/s,the length of the tube must be

Accepted Answer

For a tube closed at one end, the frequency of the nth harmonic is given by:
f_n = n*v/(4*L)
where v is the speed of sound and L is the length of the tube.

For the third harmonic, n = 3 and f_3 = 735 Hz. Substituting these values and solving for L:
L = n*v/(4*f_n)
L = 3*335/(4*735)
L = 34.1 cm

Therefore, the correct answer is option C, 34.1 cm.

Question 3

A clarinet,which is essentially a tube that is open at one end,is properly tuned to concert A (440 Hz)indoors,where the temperature is 20ºC and the speed of sound is 340 m/s.The musician then takes the instrument to play an outdoor concert,where the temperature is 0ºC and the speed of sound is 331 m/s.What is the frequency of the A played on the cold clarinet? (Ignore any thermal changes in the body of the clarinet itself.)

Accepted Answer

The frequency of the sound produced by the clarinet is proportional to the speed of sound. Since the speed of sound decreases from 340 m/s to 331 m/s, the frequency will also decrease. The new frequency can be calculated using the ratio of the speeds: (331/340) * 440 Hz = 428 Hz.

Question 4

Standing waves exist in a string of length L that is fixed at one end and free at the other.The speed of the waves on the string is v.The three lowest frequencies of vibration are

Accepted Answer

For a string fixed at one end and free at the other, the allowed modes of vibration correspond to odd multiples of a quarter wavelength fitting into the length of the string. This results in frequencies of v/4L, 3v/4L, and 5v/4L for the three lowest modes.

Question 5

For a tube of length 57.0 cm that is open at both ends,what is the frequency of the fundamental mode? (the speed of sound in air is 340 m/s)

Accepted Answer

The fundamental frequency for a tube 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 tube. Substituting $ v = 340 \, \text{m/s} $ and $ L = 0.57 \, \text{m} $, we get $ f = \frac{340}{2 \times 0.57} \approx 298 \, \text{Hz} $.

Question 6

A 1.00 m string fixed at both ends vibrates in its fundamental mode at 440 Hz.What is the speed of the waves on this string?

Accepted Answer

The speed of the waves on the string can be calculated using the formula for the fundamental frequency of a string fixed at both ends: $f = \frac{v}{2L}$, where $f$ is the frequency, $v$ is the speed of the wave, and $L$ is the length of the string. Rearranging for $v$, we get $v = 2Lf$. Substituting $L = 1.00 m$ and $f = 440 Hz$, we get $v = 2 \times 1.00 m \times 440 Hz = 880 m/s$.

Question 7

The standing waves in air in a pipe of length L that is open at one end and closed at the other have a speed v.The frequencies of the three lowest harmonics are

Accepted Answer

In a pipe that is open at one end and closed at the other, the harmonics are odd multiples of the fundamental frequency. The fundamental frequency corresponds to a quarter wavelength fitting in the pipe, so the frequencies are v/4L (fundamental), 3v/4L (third harmonic), and 5v/4L (fifth harmonic).

Question 8

When an organ pipe,which is closed at one end only,vibrates with a frequency that is three times its fundamental (first harmonic)frequency,

Accepted Answer

When an organ pipe, which is closed at one end only, vibrates with a frequency that is three times its fundamental frequency, it produces the third harmonic. The wavelength of the third harmonic is one-third of the wavelength of the fundamental frequency. Therefore, the sound produced has one-third its former wavelength.

Question 9

A string fixed at both ends is vibrating in a standing wave.There are three nodes between the ends of the string,not including those on the ends.The string is vibrating at a frequency that is its

Accepted Answer

The presence of three nodes between the ends indicates four segments or loops, which corresponds to the fourth harmonic.

Question 10

A string fixed at both ends is 50.0 cm long and has a tension that causes the frequency of its fundamental to be 262 Hz.If the tension is increased by 4%,what does the fundamental frequency become?

Accepted Answer

The frequency of the fundamental mode of a string fixed at both ends is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length of the string, $T$ is the tension in the string, and $\mu$ is the mass per unit length of the string. When the tension is increased by 4%, the new tension $T'$ is $1.04T$. The new frequency $f'$ can be found by substituting $T'$ into the formula, resulting in $f' = \frac{1}{2L} \sqrt{\frac{1.04T}{\mu}} = f \sqrt{1.04}$. Since the original frequency $f$ is 262 Hz, $f' = 262 \sqrt{1.04}$, which calculates to approximately 267 Hz.

Question 11

On a standing-wave pattern,the distance between two consecutive nodes is d.The wavelength is

Accepted Answer

In a standing wave, the distance between two consecutive nodes is half the wavelength, so the wavelength is 2d.

Question 12

The standing waves in air in a pipe of length L that is open at both ends have a speed v.The frequencies of the three lowest harmonics are

Accepted Answer

For a pipe open at both ends, the possible standing waves are those for which both ends are nodes. This means that the wavelength of the wave must be equal to twice the length of the pipe (since one half wavelength fits between each end of the pipe). 
So, we have λ = 2L for all harmonics.
The speed of sound in air is v = fλ, where f is the frequency of the wave.
For the lowest harmonic (the fundamental), we have v = f1λ = f1(2L). Solving for f1, we get f1 = v/(2L).
For the second harmonic, we have one full wavelength fitting into the length of the pipe, so we have λ = L. Thus, f2 = v/λ = v/L.
For the third harmonic, we have three half wavelengths fitting into the length of the pipe, so we have λ = 2L/3. Thus, f3 = v/λ = 3v/2L.
Therefore, the frequencies of the three lowest harmonics are v/2L, v/L, and 3v/2L, which corresponds to choice B.

Question 13

A vibrating tuning fork of frequency 1080 Hz is held above a tube filled with water.Assume the speed of sound to be 330 m/s.As the water level is lowered,consecutive maxima in intensity are observed at intervals of about

Accepted Answer

The answer of A vibrating tuning fork of frequency 1080...

Question 14

Sound has a velocity of 335 m/s in air.For an air column that is closed at both ends to resonate to a frequency of 528 Hz,the length of the air column could be

Accepted Answer

The answer of Sound has a velocity of 335 m/s...

Question 15

The standing waves on a string of length L that is fixed at both ends have a speed v.The three lowest frequencies of vibration are

Accepted Answer

The frequencies of the standing waves on a string of length L fixed at both ends are given by f = nv/2L, where n is a positive integer. The three lowest frequencies are obtained for n = 1, 2, and 3.
Substituting n = 1, 2, and 3, we get:
f1 = v/2L
f2 = 2v/2L = v/L
f3 = 3v/2L
Hence, option B is correct.

Question 16

The human vocal tract can be thought of as a tube that is open at one end.If the length of this tube is 17 cm (about average for an adult male),what are the lowest two harmonics?

Accepted Answer

The human vocal tract can be modeled as a tube open at one end, which means it supports odd harmonics. The fundamental frequency (first harmonic) can be found using the formula $f = \frac{v}{4L}$, where $v$ is the speed of sound (approximately 340 m/s) and $L$ is the length of the tube. For a 17 cm (0.17 m) tube, $f = \frac{340}{4 \times 0.17} \approx 500 \, \text{Hz}$. The next harmonic in a tube open at one end is the third harmonic, which is three times the fundamental frequency, thus $3 \times 500 \, \text{Hz} = 1500 \, \text{Hz}$.

Question 17

The ratio of the fundamental frequency (first harmonic)of an open pipe to that of a closed pipe of the same length is

Accepted Answer

The fundamental frequency of an open pipe is given by $f = \frac{v}{2L}$, where $v$ is the speed of sound and $L$ is the length of the pipe. For a closed pipe, the fundamental frequency is $f = \frac{v}{4L}$. Therefore, the ratio of the fundamental frequency of an open pipe to that of a closed pipe is $\frac{\frac{v}{2L}}{\frac{v}{4L}} = 2:1$.

Question 18

In a pipe that is open at one end and closed at the other and that has a fundamental frequency of 256 Hz,which of the following frequencies cannot be produced?

Accepted Answer

The answer of In a pipe that is open at...

Question 19

A stretched string of length L,fixed at both ends,is vibrating in its third harmonic.How far from the end of the string can the blade of a screwdriver be placed against the string without disturbing the amplitude of the vibration?

Accepted Answer

The answer of A stretched string of length L,fixed at...

Question 20

A vibrating tuning fork of frequency 640 Hz is held above a tube filled with water.Assume the speed of sound to be 330 m/s.As the water level is lowered,consecutive maxima in intensity are observed at intervals of about

Accepted Answer

The wavelength of sound in the tube is equal to four times the length of the tube when a resonance occurs. Therefore,
λ = 4L

Speed of sound, v = 330 m/s
Frequency of tuning fork, f = 640 Hz

The equation of speed of sound, frequency, and wavelength is given by:
v = fλ

Therefore,
λ = v/f = 330/640 = 0.51562 m

One-fourth of wavelength for the resonance to occur is equal to the length of the air column. Therefore,
L = λ/4 = 0.51562/4 = 0.1289 m = 12.9 cm

Hence, consecutive maxima in intensity are observed at intervals equal to the length of the air column in the tube, which is 12.9 cm. Thus, the correct choice is C.

Quiz 16: Superposition and Standing Waves