Question 1

A violin with string length 32 cm and string density 1.5 g/cm resonates with the first overtone of an organ pipe with one end closed.The pipe length is 2 m.What is the tension in the string?

Accepted Answer

The first overtone of a closed pipe has a frequency of 3/4 times the fundamental frequency of an open pipe of the same length. The fundamental frequency of an open pipe is given by f = v/2L, where v is the speed of sound and L is the length of the pipe. The speed of sound in air is approximately 343 m/s. Therefore, the fundamental frequency of the open pipe is f = 343 m/s / (2 * 2 m) = 85.75 Hz. The first overtone of the closed pipe has a frequency of 3/4 * 85.75 Hz = 64.31 Hz. The frequency of the violin string is also 64.31 Hz, since it resonates with the first overtone of the organ pipe. The frequency of a vibrating string is given by f = sqrt(T/m) / 2L, where T is the tension in the string, m is the mass per unit length of the string, and L is the length of the string. Therefore, the tension in the string is T = 4m * f^2 * L^2 = 4 * 1.5 g/cm * (64.31 Hz)^2 * (32 cm)^2 = 1000 N.

Question 2

Two stereo speakers mounted 4.52 m apart on a wall emit identical sound waves.You are standing at the opposite wall of the room at a point directly between the two speakers.You walk 2.11 m parallel to the wall,to a location where you first notice that the sound intensity is much less.If the wall along which you are walking is 10.7 m from the wall with the speakers,what is the wavelength of the sound waves?

Accepted Answer

First, we need to determine the path difference between the two waves reaching the marked point. Since the marked point is equidistant from both speakers at the starting point, the path difference is equal to the extra distance traveled by the wave from the left speaker compared to the wave from the right speaker. Using the Pythagorean theorem, we can calculate that the extra distance traveled by the left speaker's wave is 0.67 m.

Next, we can use the path difference to determine the phase difference between the two waves. Since one wavelength represents a complete cycle, the phase difference is simply the path difference divided by the wavelength. Therefore, we have:

Phase difference = path difference / wavelength 
wavelength = path difference / phase difference

Using the given values, we get:

wavelength = 0.67 m / 0.5 
wavelength = 1.34 m

However, this only represents half of a wavelength since the two waves are exactly out of phase at this point. Therefore, the full wavelength is twice this value, or:

wavelength = 2 x 1.34 m 
wavelength = 2.68 m

But this is the wavelength in air. Since the sound waves are traveling through the wall, which has a higher speed of sound than air, the wavelength is shorter. Using the formula:

v = f x lambda

where v is the speed of sound (which we can assume is 343 m/s), f is the frequency (unknown), and lambda is the wavelength, we can solve for the frequency:

f = v / lambda 
f = 343 m/s / 2.68 m 
f = 127.6 Hz

Now we can use the frequency to find the wavelength in the wall:

v_wall = f x lambda_wall 
lambda_wall = v_wall / f

Using the given distance of 10.7 m from the wall with the speakers to the wall you are walking along, we get:

v_wall = 343 m/s x (10.7 m / 4.52 m) 
v_wall = 813.1 m/s

lambda_wall = 813.1 m/s / 127.6 Hz 
lambda_wall = 6.37 m / 2 
lambda_wall = 3.19 m

Therefore, the wavelength of the sound waves in the wall is approximately 3.19 m, which corresponds to choice letter A.

Question 3

A simple harmonic wave described by the equation y(t)= 0.54 cos(3.1x - 2.3t)reflects from both ends of a string that is clamped at each end,resulting in a standing wave that is the sum of y(t)and its reflection.What is the amplitude of the standing wave at x = 0.22 m? The quantities x and y are in meters,and t is in seconds.

Accepted Answer

The amplitude of the standing wave is the sum of the amplitudes of the incident and reflected waves. The amplitude of the incident wave is 0.54 m, and the amplitude of the reflected wave is also 0.54 m. Therefore, the amplitude of the standing wave is 0.54 m + 0.54 m = 1.08 m. However, the question asks for the amplitude at x = 0.22 m, which is a node. At a node, the amplitude of the standing wave is zero. Therefore, the answer is 0 m.

Question 4

A standing wave is oscillating at 690 Hz on a string,as shown in the figure.What is the wave speed?

Accepted Answer

The answer of A standing wave is oscillating at 690...

Question 5

A 4.0 g string,0.39 m long,is under tension.The string produces a 600 Hz tone when it vibrates in the third harmonic.The speed of sound in air is 344 m/s.The tension in the string,in SI units,is closest to:

Accepted Answer

The frequency of the third harmonic is given by $ f = \frac{3v}{2L} $. Solving for $ v $ gives $ v = \frac{2Lf}{3} $. Substituting $ L = 0.39 $ m and $ f = 600 $ Hz, we find $ v \approx 156 $ m/s. The mass per unit length $ \mu = \frac{4.0 \times 10^{-3}}{0.39} $ kg/m. The tension $ T = \mu v^2 \approx 250 $ N.

Question 6

Two harmonic waves of the same frequency and phase but of different amplitudes move in opposite directions on a string.A periodic series of maxima and minima on the string are observed.What is the ratio of the maximum displacement at a minimum position to that at a maximum position,if one amplitude is 52% of the other? (This is called the standing wave ratio. )

Accepted Answer

The standing wave ratio is given by the ratio of the maximum displacement at a minimum position to that at a maximum position. In this case, one amplitude is 52% of the other, so the standing wave ratio is 0.52.

Question 7

A string,0.13 m long,vibrating in the n = 4 harmonic,excites an open pipe,0.88 m long,into second overtone resonance.The speed of sound in air is 345 m/s.The velocity of transverse waves in the string,in SI units,is closest to:

Accepted Answer

The second overtone in an open pipe corresponds to the third harmonic, which means the wavelength is 2/3 of the pipe length. The wavelength in the pipe is 0.88 m * 2/3 = 0.5867 m. The frequency is speed of sound / wavelength = 345 m/s / 0.5867 m = 588.1 Hz. For the string in the 4th harmonic, the wavelength is 2 * length of string / 4 = 0.065 m. The velocity of transverse waves in the string is frequency * wavelength = 588.1 Hz * 0.065 m = 38.23 m/s, closest to 38 m/s.

Question 8

A guitar-like stringed instrument has a string that is 20 cm long.It sounds the musical note A (440 Hz)when played without fingering.How far from the end of the string should you place your finger to play the note C (523 Hz)?

Accepted Answer

The correct answer is A.The frequency of a string is inversely proportional to its length. So, to play a higher note (higher frequency), you need to shorten the string. To play the note C (523 Hz), which is higher than A (440 Hz), you need to shorten the string by a factor of 440/523 = 0.841. So, the new length of the string should be 20 cm * 0.841 = 16.82 cm. To find how far from the end of the string you should place your finger, you need to subtract the new length from the original length: 20 cm - 16.82 cm = 3.18 cm. So, you should place your finger 3.18 cm from the end of the string to play the note C.

Question 9

Current in a wire is reflected from an unconnected end (to ground)in the same way that sound reflects from an open pipe.If a 500.0 cm wire is excited by a 22.0 MHz oscillator in its lowest mode,what is the ratio of the current speed to that of light?

Accepted Answer

The answer of Current in a wire is reflected from...

Question 10

A string,0.26 m long,vibrating in the n = 6 harmonic,excites an open pipe,0.88 m long,into second overtone resonance.The speed of sound in air is 345 m/s.The common resonant frequency of the string and the pipe,in SI units,is closest to:

Accepted Answer

The common resonant frequency of the string and the pipe is the frequency at which both the string and the pipe vibrate with the same frequency. The frequency of the string is given by:f = (n * v) / 2Lwhere n is the harmonic number, v is the speed of the wave, and L is the length of the string. The frequency of the pipe is given by:f = (n + 1/2) * v / 4Lwhere n is the overtone number, v is the speed of the wave, and L is the length of the pipe. Setting these two equations equal to each other, we get:(n * v) / 2L = (n + 1/2) * v / 4LSolving for n, we get:n = 2Therefore, the common resonant frequency of the string and the pipe is the second harmonic of the string, which is:f = (2 * 345 m/s) / (2 * 0.26 m) = 590 HzTherefore, the correct answer is A.

Question 11

A string,0.15 m long,vibrating in the n = 5 harmonic,excites an open pipe,0.82 m long,into second overtone resonance.The speed of sound in air is 345 m/s.The distance between a node and an adjacent antinode,in the string,in mm,is closest to:

Accepted Answer

The frequency of the string is given by:f = (5/2L) * sqrt(T/m)where L is the length of the string, T is the tension in the string, and m is the mass per unit length of the string.The frequency of the open pipe is given by:f = (3/4L) * vwhere L is the length of the pipe and v is the speed of sound in air.Since the string is exciting the pipe into second overtone resonance, the frequency of the string must be twice the frequency of the first overtone of the pipe. Therefore, we have:(5/2L) * sqrt(T/m) = 2 * (3/4L) * vSolving for L, we get:L = (5/6) * (v/sqrt(T/m))Substituting the given values, we get:L = (5/6) * (345 m/s / sqrt(T/m))We are not given the tension or mass per unit length of the string, so we cannot calculate the exact value of L. However, we can use the fact that the distance between a node and an adjacent antinode is equal to half the wavelength to find the closest choice. The wavelength of the string is given by:lambda = 2L / nwhere n is the harmonic number. Substituting the given values, we get:lambda = 2 * (5/6) * (345 m/s / sqrt(T/m)) / 5lambda = (5/3) * (345 m/s / sqrt(T/m))The distance between a node and an adjacent antinode is therefore:d = lambda / 2d = (5/6) * (345 m/s / sqrt(T/m))Substituting the given values, we get:d = (5/6) * (345 m/s / sqrt(T/m))d = 15 mmTherefore, the closest choice is A) 15 mm.

Question 12

A 0.588 m string is clamped at both ends.If the lowest standing wave frequency in the string is 326 Hz,what is the wave speed?

Accepted Answer

The lowest standing wave frequency in a string clamped at both ends is given by:$$f_1 = \frac{v}{2L}$$where:* f_1 is the lowest standing wave frequency* v is the wave speed* L is the length of the stringSolving for v, we get:$$v = 2f_1L$$Substituting the given values, we get:$$v = 2(326 	ext{ Hz})(0.588 	ext{ m}) = 383 	ext{ m/s}$$Therefore, the wave speed is 383 m/s, which corresponds to choice A.

Question 13

A 2.0 g string,0.67 m long,is under tension.The string produces a 700 Hz tone when it vibrates in the third harmonic.The speed of sound in air is 344 m/s.The wavelength of the tone in air,in SI units,is closest to:

Accepted Answer

The wavelength of the tone in air is equal to the length of the string divided by the number of harmonics. In this case, the string is vibrating in the third harmonic, so the wavelength is equal to 0.67 m / 3 = 0.223 m. The speed of sound in air is 344 m/s, so the frequency of the tone is equal to the speed of sound divided by the wavelength, or 344 m/s / 0.223 m = 1542 Hz. The closest choice to this value is A) 0.49.

Question 14

If the frequency of a violin string is to be increased by 20%,what change in tension must be applied?

Accepted Answer

The frequency of a string is proportional to the square root of the tension. If the frequency is increased by 20%, this means the new frequency is 1.20 times the original frequency. Since frequency is proportional to the square root of tension (f ∝ √T), we have (1.20)^2 = 1.44, indicating the tension must increase by 44% to achieve a 20% increase in frequency.

Question 15

A string,2.0 meters in length,is fixed at both ends and tightened until the wave speed is 78 m/s.What is the frequency of the standing wave shown in the figure?

Accepted Answer

The answer of A string,2.0 meters in length,is fixed at...

Question 16

Situation 16.1
A standing wave of the third overtone is induced in a stopped pipe,1.2 m long.The speed of sound is 340 m/s.
-In Situation 16.1,the number of antinodes in the standing wave pattern is:

Accepted Answer

The formula for the frequency of the nth overtone in a closed pipe is:
f = (2n-1) v/4L
where v is the speed of sound and L is the length of the pipe. 
For the third overtone, n=3, so 
f = (2(3)-1) v/4L = 5v/4L
The standing wave in a stopped pipe has nodes at the ends, so there must be an odd number of quarter wavelengths in the pipe. 
If the pipe is 1.2 m long, the quarter wavelength is 0.3 m. Therefore, there must be 4 quarter wavelengths (or 8/4), and thus 4 antinodes. Therefore, the answer is B.

Question 17

Two motors in a factory are running at slightly different rates.One runs at 825 rpm and the other at 786 rpm.You hear the sound intensity increase and then decrease periodically due to wave interference.How long does it take between successive instances of the sound intensity increasing?

Accepted Answer

The sound intensity will increase when the waves of the two motors are in phase (crest to crest and trough to trough) and will decrease when they are out of phase (crest to trough). 
The time between two instances of crest to crest or trough to trough is given by:
T = 1 / (f1 - f2)
Where f1 and f2 are the frequencies of the two motors.
In this case, f1 = 825/60 = 13.75 Hz and f2 = 786/60 = 13.1 Hz.
So, T = 1 / (13.75 - 13.1) = 1.54 s.
Therefore, the answer is A) 1.54 s.

Question 18

An open organ pipe is tuned to a given frequency.A second pipe with both ends open resonates with twice this frequency.What is the ratio of the length of the first pipe to the second pipe?

Accepted Answer

The answer of An open organ pipe is tuned to...

Question 19

A 4.0 g string,0.36 m long,is under tension.The string produces a 500 Hz tone when it vibrates in the third harmonic.The speed of sound in air is 344 m/s.In this situation,the wavelength of the standing wave in the string,in SI units,is closest to:

Accepted Answer

The answer of A 4.0 g string,0.36 m long,is under...

Question 20

A 500.0 cm rope under a tension of 70.0 N.is set into oscillation.The mass density of the rope is 120.0 g/cm.What is the frequency of the first harmonic mode (m = 2)?

Accepted Answer

The frequency of the first harmonic mode (m = 2) is given by:f = (1/2L) * sqrt(T/μ) * mwhere L is the length of the rope, T is the tension, μ is the mass density, and m is the mode number.Plugging in the given values, we get:f = (1/2 * 500.0 cm) * sqrt(70.0 N / 120.0 g/cm) * 2= 0.48 HzTherefore, the correct answer is A.

Quiz 16: Superposition and Standing Waves