Question 1

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
A) 11.6 cm
B) 22.9 cm
C) 34.1 cm
D) 45.7 cm
E) 57.3 cm

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 2

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
A) 64.0 Hz
B) 128 Hz
C) 256 Hz
D) 512 Hz
E) 1.02 kHz

Accepted Answer

The answer of The fundamental frequency of a pipe that...

Question 3

The wave function y(x,t) for a standing wave on a string fixed at both ends is given by y(x,t) = 0.080 sin 6.0x cos 600t where the units are SI. The wavelength of this wave is
A) 6.00 m
B) 1.05 m
C) 600 m
D) 0.010 m
E) impossible to tell given this information about the standing wave.

Accepted Answer

In a standing wave on a string fixed at both ends, the wavelength is given by 2L/n where L is the length of the string and n is the number of nodes (points of zero amplitude) between the ends.

In the given wave function, the standing wave is a combination of a sine and cosine function with different spatial frequencies. The sine function has a spatial frequency of 6.0 waves per unit length (presumably meters, since the units are in SI), while the cosine function has a spatial frequency of 600 waves per unit time (presumably seconds).

The wavelength of the standing wave will be determined by the lowest common multiple of these two frequencies, which is 0.0105 m or 1.05 cm. This is the distance between adjacent nodes in the wave.

Therefore, the correct answer is B) 1.05 m.

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
A) v/4L, v/2L, and 3v/4L
B) v/2L, v/L, and 3v/2L
C) $\lambda$/4, $\lambda$/2, and 3$\lambda$/4
D) v/4L, 3v/4L, and 5v/4L
E) $\lambda$/3, 2$\lambda$/3, and 3$\lambda$/3

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

A string with mass density equal to 0.0025 kg/m is fixed at both ends and at a tension of 290 N. Resonant frequencies are found at 558 Hz and the next one at 744 Hz. To what harmonic does the 558 Hz resonance correspond?
A) 1
B) 2
C) 3
D) 4
E) 5

Accepted Answer

The answer of A string with mass density equal to...

Question 6

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
A) v/L, 2v/L, and 3v/L
B) v/2L, v/L, and 3v/2L
C) $\lambda$/2, $\lambda$, and 3$\lambda$/2
D) L/v, 2L/v, and 3L/v
E) $\lambda$/3, 2$\lambda$/3, and 3$\lambda$/3

Accepted Answer

The frequency of a standing wave on a string fixed at both ends is given by f = n(v/2L), where n is an integer (1, 2, 3, etc.) and v is the speed of the wave on the string. The three lowest frequencies correspond to n = 1, 2, and 3, so the frequencies are:
f1 = v/2L
f2 = 2v/2L = v/L
f3 = 3v/2L
Choice B lists these frequencies in the correct order, so it is the best choice.

Question 7

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?
A) 220 m/s
B) 440 m/s
C) 660 m/s
D) 880 m/s
E) 1.10 km/s

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 8

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)
A) 149 Hz
B) 447 Hz
C) 596 Hz
D) 298 Hz
E) 746 Hz

Accepted Answer

The answer of For a tube of length 57.0 cm...

Question 9

The wave function y(x,t) for a standing wave on a string fixed at both ends is given by y(x,t) = 0.080 sin 6.0x cos 600t where the units are SI. The speed of the traveling waves that result in this standing wave is
A) 6.00 m
B) 1.05 m
C) 600 m
D) 0.010 m
E) impossible to tell given this information about the standing wave.

Accepted Answer

The answer of The wave function y(x,t) for a standing...

Question 10

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
A) 79.2 cm
B) 55.5 cm
C) 47.5 cm
D) 31.7 cm
E) 15.8 cm

Accepted Answer

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

Question 11

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
A) v/L, 2v/L, and 3v/L
B) v/2L, v/L, and 3v/2L
C) $\lambda$/2, $\lambda$, and 3$\lambda$/2
D) L/v, 2L/v, and 3L/v
E) $\lambda$/3, 2$\lambda$/3, and 3$\lambda$/3

Accepted Answer

The frequencies of the standing waves in a pipe open at both ends are given by f = nv/2L, where n is the harmonic number (1, 2, 3, etc.), v is the speed of sound in the air, and L is the length of the pipe. Plugging in n = 1, 2, and 3, we get f = v/2L, v/L, and 3v/2L, respectively. Therefore, the correct answer is (B) v/2L, v/L, and 3v/2L.

Question 12

A string with mass density equal to 0.0025 kg/m is fixed at both ends and at a tension of 290 N. Resonant frequencies are found at 558 Hz and the next one at 744 Hz. What is the fundamental frequency of the string?
A) 558 Hz
B) 372 Hz
C) 93 Hz
D) 186 Hz
E) none of the above

Accepted Answer

The resonant frequencies of a string fixed at both ends are given by:

fn = n(v/2L)

where fn is the nth resonant frequency, v is the speed of sound in the string, L is the length of the string, and n is a positive integer.

In this case, we are given two resonant frequencies: 558 Hz and 744 Hz. We can set up two equations using the formula above and solve for the speed of sound v:

558 = (1)(v/2L)
744 = (2)(v/2L)

Dividing the second equation by the first, we get:

744/558 = 2/1 = 4/2

Simplifying, we get:

4/3 = (v/2L)/(v/2L) = 1

This means the ratio of the lengths of the string for the two resonant frequencies is 4/3.

We can use this ratio to find the fundamental frequency f1. The fundamental frequency is the first resonant frequency, n = 1, so we can use the formula above with n = 1:

f1 = (v/2L)

We want to find f1 in terms of the given resonant frequencies, so we can use the ratio:

4/3 = L2/L1

Solving for L1, we get:

L1 = (3/4)L2

Substituting into the formula for f1, we get:

f1 = (v/2L1) = (v/2(3/4)L2) = (2/3)(v/2L2) = (2/3)(558) = 372 Hz

Therefore, the answer is D) 186 Hz, since it is not one of the given choices.

Question 13

The wave function y(x,t) for a standing wave on a string fixed at both ends is given by y(x,t) = 0.080 sin 6.0x cos 600t where the units are SI. The distance between successive nodes on the string is
A) 0.24 m
B) 0.08 m
C) 0.02 m
D) 0.52 m
E) impossible to tell given this information about the standing wave.

Accepted Answer

The answer of The wave function y(x,t) for a standing...

Question 14

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
A) v/4L, v/2L, and 3v/4L
B) v/2L, v/L, and 3v/2L
C) $\lambda$/4, $\lambda$/2, and 3$\lambda$/4
D) v/4L, 3v/4L, and 5v/4L
E) $\lambda$/3, 2$\lambda$/3, and 3$\lambda$/3

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 15

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?
A) 500 Hz, 1500 Hz
B) 500 Hz, 1000 Hz
C) 1000 Hz, 2000 Hz
D) 1000 Hz, 3000 Hz
E) 1500 Hz, 2500 Hz

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 16

The ratio of the fundamental frequency (first harmonic) of an open pipe to that of a closed pipe of the same length is
A) 2:1
B) 7:8
C) 4:5
D) 3:2
E) 1:2

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 17

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?
A) 252 Hz
B) 257 Hz
C) 264 Hz
D) 267 Hz
E) 272 Hz

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 18

The sound wave in an organ tube has a wavelength that is equal to the distance between
A) A and B.
B) A and C.
C) the nodes farthest apart.
D) the antinodes farthest apart.
E) None of these is correct.

Accepted Answer

The answer of  The sound wave in an organ...

Question 19

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.)
A) 417 Hz
B) 428 Hz
C) 434 Hz
D) 445 Hz
E) 451 Hz

Accepted Answer

The answer of A clarinet, which is essentially a tube...

Question 20

The wave function y(x,t) for a standing wave on a string fixed at both ends is given by y(x,t) = 0.080 sin 6.0x cos 600t where the units are SI. The amplitudes of the traveling wave that result in this standing wave are
A) 0.04 m
B) 0.08 m
C) 0.02 m
D) 0.16 m
E) impossible to tell given this information about the standing wave.

Accepted Answer

For a standing wave on a string fixed at both ends, the amplitude of the traveling wave can be calculated using the following equation:
Amplitude of traveling wave = (1/2) * Amplitude of standing wave

From the given equation:
y(x,t) = 0.080 sin 6.0x cos 600t

The amplitude of the standing wave is 0.080 m.

Therefore, the amplitude of the traveling wave is (1/2) * 0.080 m = 0.04 m.

Hence, the best choice is A) 0.04 m.

Quiz 16: Superposition and Standing Waves

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

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

The wave function y(x,t) for a standing wave on a string fixed at both ends is given by y(x,t) = 0.080 sin 6.0x cos 600t where the units are SI. The wavelength of this wave is

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

A string with mass density equal to 0.0025 kg/m is fixed at both ends and at a tension of 290 N. Resonant frequencies are found at 558 Hz and the next one at 744 Hz. To what harmonic does the 558 Hz resonance correspond?

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

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?

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)

The wave function y(x,t) for a standing wave on a string fixed at both ends is given by y(x,t) = 0.080 sin 6.0x cos 600t where the units are SI. The speed of the traveling waves that result in this standing wave is

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

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

A string with mass density equal to 0.0025 kg/m is fixed at both ends and at a tension of 290 N. Resonant frequencies are found at 558 Hz and the next one at 744 Hz. What is the fundamental frequency of the string?

The wave function y(x,t) for a standing wave on a string fixed at both ends is given by y(x,t) = 0.080 sin 6.0x cos 600t where the units are SI. The distance between successive nodes on the string is

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

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?

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

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?

The sound wave in an organ tube has a wavelength that is equal to the distance between

The wave function y(x,t) for a standing wave on a string fixed at both ends is given by y(x,t) = 0.080 sin 6.0x cos 600t where the units are SI. The amplitudes of the traveling wave that result in this standing wave are