Question 1

The reason we can tell the difference between a trumpet and a clarinet when they both play the same pitch is that they have
A)the same overtones.
B)the same harmonics.
C)different fundamental frequencies.
D)different waveforms.
E)harmonic syntheses.

Accepted Answer

The waveform of a sound is its shape over time and is determined by the specific combination and strength of its harmonics. Different instruments have different waveforms due to the unique combination of harmonics they produce, allowing us to distinguish between them even when playing the same pitch. Therefore, the correct choice is D, different waveforms.

Question 2

Wire A is the same mass per unit length as wire B.However wire A is twice as long as wire B and has three times as much tension on it.Calculate the fundamental frequency of wire A divided by wire B.
A)0.87
B)0.66
C)0.43
D)0.75
E)1.50

Accepted Answer

The fundamental frequency of a wire is given by:$$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$where:* f is the fundamental frequency* L is the length of the wire* T is the tension on the wire* $\mu$ is the mass per unit length of the wireSince wire A has the same mass per unit length as wire B, $\mu$ is the same for both wires. Since wire A is twice as long as wire B, $L_A = 2L_B$. Since wire A has three times as much tension on it, $T_A = 3T_B$. Substituting these values into the equation for the fundamental frequency, we get:$$f_A = \frac{1}{2(2L_B)}\sqrt{\frac{3T_B}{\mu}} = \frac{1}{4L_B}\sqrt{\frac{3T_B}{\mu}}$$$$f_B = \frac{1}{2L_B}\sqrt{\frac{T_B}{\mu}}$$Dividing $f_A$ by $f_B$, we get:$$\frac{f_A}{f_B} = \frac{\frac{1}{4L_B}\sqrt{\frac{3T_B}{\mu}}}{\frac{1}{2L_B}\sqrt{\frac{T_B}{\mu}}} = \frac{1}{2}\sqrt{\frac{3}{1}} = \frac{1}{2}\sqrt{3} = 0.866$$Therefore, the fundamental frequency of wire A divided by wire B is 0.866, which is closest to choice A, 0.87.

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 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.

Question 4

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 wave...

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.What is the length of the wire?
A)0.8 m
B)1.6 m
C)3.2 m
D)1.2 m
E)0.4 m

Accepted Answer

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

Question 6

Two pipes closed at one end of length L₁ and L₂ are excited at their resonant frequencies.If the beat period is B_f Hz,then the velocity of sound is given by: A)B_f $\times$ L₁ $\times$ L₂ / (4L₁ $\times$ 4L₂) B)4 $\times$ B_f $\times$ L₁ $\times$ L₂ / (4L₁ - 4L₂) C)16 $\times$ B_f $\times$ L₁ $\times$ L₂ / (4L₁ + 4L₂) D)4 $\times$ B_f $\times$ L₁ $\times$ L₂ / (L₁ -L₂) E)4 $\times$ B_f $\times$ L₁ $\times$ L₂ / (L₁ + L₂)

Accepted Answer

The answer of Two pipes closed at one end of...

Question 7

A pipe produces successive harmonics at 300 Hz and 350 Hz.Calculate the length of the pipe and state whether it is closed at one end or not.Assume the speed of sound to be 340 m/s.
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of A pipe produces successive harmonics at 300...

Question 8

A vibrating tuning fork of 725 Hz is held above a tube filled with water.Successive resonances are heard when the water level is lowered by 11.5 cm and 34.5 cm from the top of the tube.Calculate a value for the speed of sound.(Hint: remember the small end correction $\Delta$L at the top of the tube.)
A)333 m/s
B)343 m/s
C)325 m/s
D)315 m/s
E)338 m/s

Accepted Answer

The correct answer is A because the speed of sound is 333 m/s.

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 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 fixed at both ends, the boundary conditions require that the ends of the wave must be nodes. Therefore, the wavelength of the standing wave is twice the distance between the ends, or:
λ = 2L
where L is the length of the string.
In this problem, we are not given the length of the string, but we can use the fact that the standing wave has six nodes (points of zero displacement) between the two ends. The distance between adjacent nodes is λ/2, so the length of the string is:
L = (3/2)λ
We can now solve for the wavelength:
λ = 2L/3 = 2/3(3/2) = 1.0 m
Therefore, the best choice is B) 1.05 m.

Question 10

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 wave...

Question 11

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 difference between consecutive resonant frequencies is the fundamental frequency. The difference between 744 Hz and 558 Hz is 186 Hz, which is the fundamental frequency. The 558 Hz resonance is the third harmonic (558 Hz / 186 Hz = 3).

Question 12

A piano tuner hears a beat every 0.33 seconds when he hits a note and compares it to his reference tone at 163 Hz.What is the lowest possible frequency of the piano note?
A)44.9 Hz
B)166.0 Hz
C)162.7 Hz
D)163.3 Hz
E)160.0 Hz

Accepted Answer

The answer of A piano tuner hears a beat every...

Question 13

A wire of mass 1.1 g is under a tension of 100 N.If its third overtone is at a frequency of 750 Hz,calculate the length of the wire.
A)72 cm
B)101 cm
C)36 cm
D)65 cm
E)None of the above

Accepted Answer

The answer of A wire of mass 1.1 g is...

Question 14

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 n is the harmonic number (n=1 for fundamental frequency), v is the velocity of the wave on the string, and L is the length of the string.

The velocity of the wave on the string can be found using the tension and mass density of the string:
v = √(T/μ),

where T is the tension and μ is the mass density.

Substituting the given values, we get:
v = √(290/0.0025) = 240.5 m/s

Using the resonant frequencies given, we can write two equations:
558 = (1/2L)(240.5)
744 = (3/2L)(240.5)

Solving these equations simultaneously gives:
L = 0.690 m

Using the formula for the fundamental frequency, we get:
f1 = (1/2L)(240.5) = 173.6 Hz

Therefore, the correct answer is D) 186 Hz (rounded to the nearest whole number).

Question 15

A vibrating tuning fork of 850 Hz is held above a tube filled with water.The first and third resonances occur when the water level is lowered by 8.8 cm and 47.6 cm from the top of the tube.If there is a small end correction that adds a small extra length $\Delta$L to the effective length of the air column,calculate $\Delta$L.Assume the speed of sound to be 330 m/s.
A)0.2 cm
B)0.9 cm
C)0.4 cm
D)0.6 cm
E)1.1 cm

Accepted Answer

The answer of A vibrating tuning fork of 850 Hz...

Question 16

Two identical loudspeakers are driven in phase by the same amplifier.The speakers are positioned a distance of 3.2 m apart.A person stands 4.1 m away from one speaker and 4.8 m away from the other.Calculate the second lowest frequency that results in destructive interference at the point where the person is standing.Assume the speed of sound to be 340 m/s.
A)245 Hz
B)735 Hz
C)1225 Hz
D)490 Hz
E)1470 Hz

Accepted Answer

The path difference between the two speakers at the point where the person is standing is given by:
Δx = x2 - x1 = 4.8 - 4.1 = 0.7 m

For destructive interference, the path difference must be an odd multiple of half-wavelength, i.e., Δx = (2n + 1)λ/2, where n is an integer.

Rearranging the above equation, we get:
λ = 2(Δx)/(2n + 1)

Substituting the given values, we get:
λ = 2(0.7)/(2n + 1) = 1.4/(2n + 1)

For the second lowest frequency, n = 1, since we want the second odd multiple of half-wavelength. Hence,
f = v/λ = v(2n + 1)/2Δx = 340(2(1) + 1)/2(0.7) = 735 Hz

Therefore, the answer is B.

Question 17

A vibrating tuning fork is held above a tube filled with water.The first two resonances occur when the water level is lowered by 14.2 cm and 44.2 cm from the top of the tube.If there is a small end correction that adds a small extra length $\Delta$L? to the effective length of the air column,calculate the frequency of the tuning fork.Assume the speed of sound to be 330 m/s.
A)560 Hz
B)581 Hz
C)550 Hz
D)1100 Hz
E)1120 Hz

Accepted Answer

The answer of A vibrating tuning fork is held above...

Question 18

What is the third harmonic of an open-both-ends organ pipe of length 1.5 m? Assume the speed of sound to be 340 m/s.
A)229 Hz
B)340 Hz
C)457 Hz
D)686 Hz
E)none of the above

Accepted Answer

The answer of What is the third harmonic of an...

Question 19

A vibrating tuning fork of 300 Hz is held above a tube filled with water.The first resonance is heard when the water level is lowered by 26.1 cm.A second tuning fork of 400 Hz is held above the tube,and its first resonance occurs when the water level is lowered by 19.3 cm from the top.Calculate a value for the speed of sound.(Hint: remember the small end correction $\Delta$L at the top of the tube.)
A)333 m/s
B)343 m/s
C)325 m/s
D)315 m/s
E)338 m/s

Accepted Answer

The answer of A vibrating tuning fork of 300 Hz...

Question 20

A guitar string of length 105 cm is in resonance with a tuning fork of frequency f.Using the fret board the length of the string is shortened by 1.5 cm while keeping the tension in the string constant.Now a beat frequency of 10 Hz is heard between the string and the tuning fork.What is the frequency of the tuning fork?
A)230 Hz
B)1380 Hz
C)345 Hz
D)690 Hz
E)none of the above

Accepted Answer

The answer of A guitar string of length 105 cm...

Quiz 16: Superposition and Standing Waves

The reason we can tell the difference between a trumpet and a clarinet when they both play the same pitch is that they have

Wire A is the same mass per unit length as wire B.However wire A is twice as long as wire B and has three times as much tension on it.Calculate the fundamental frequency of wire A divided by wire B.

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

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 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 length of the wire?

Two pipes closed at one end of length L₁ and L₂ are excited at their resonant frequencies.If the beat period is B_f Hz,then the velocity of sound is given by:

A pipe produces successive harmonics at 300 Hz and 350 Hz.Calculate the length of the pipe and state whether it is closed at one end or not.Assume the speed of sound to be 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 wavelength of this wave 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 distance between successive nodes on the string is

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 piano tuner hears a beat every 0.33 seconds when he hits a note and compares it to his reference tone at 163 Hz.What is the lowest possible frequency of the piano note?

A wire of mass 1.1 g is under a tension of 100 N.If its third overtone is at a frequency of 750 Hz,calculate the length of the wire.

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?

What is the third harmonic of an open-both-ends organ pipe of length 1.5 m? Assume the speed of sound to be 340 m/s.