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In: Physics

A musician in a concert hall is tuning her wind instrument. When she plays a short...

A musician in a concert hall is tuning her wind instrument. When she plays a short note she hears the echo of the note return from the opposite side of the 50.0 meter long auditorium 0.294 seconds later. Model the instrument as a tube closed at one end, if the instrument is properly tuned the note of the musician played would have a frequency of 233.082 Hz, but instead has a frequency of 226.513 Hz. This note is the first overtone, since the fundamental frequency is suppressed on this instrument. 1) What is the speed of sound in the concert hall on that day? 2) What is the temperature of the concert hall on that day in Kelvin? 3) What is the length of the tubing on the musician's instrument in meters? 4) How much does the musician need to change the length of the tubing to be in tune on that day? Does the tube need to be shortened or lengthened? Find the change in length of the tube, and make sure to use a negative number if the tube needs to be shorter. ( answer in meters.) 5) What is the fundamental frequency of this instrument when correctly tuned? 6) What is the second overtone of the correctly tuned instrument?  

Solutions

Expert Solution

(1) speed of sound = distance / time = 50 /(0.294/2) = 340 m/s

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(2) Temperature variation speed of sound, v(T) = v0 [ (273+T)/273 ]1/2

where vo = 331.5 m/s is speed of sound at 0oC and T is in oC

Hence we have, 340 = 331.5 [ (273+T)/273 ]1/2

we get temperature of concert hall from above eqn. as, T = 14.2 oC

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(3) wavelength of untuned note = speed of sound/ frequency = 340 / 226.513 1.5 m

Tube length for first overtone of untuned note = (3/4) = 1.125 m

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(4) wavelength of tuned note= speed of sound/ frequency = 340 / 233.082 1.459 m

Proper tube length for first overtone of tuned note= (3/4) = 1.094 m

Length to be shortened = 1.125 - 1.094 = 0.031 m = 3.1 cm

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(5) fundamental frequency n3 if correctly tuned is given by, n3= 3 n1

where n1 is fundamental

hence n3 = 233.082/3 = 77.694 Hz

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(6) second overtone n5 of correctly tuned note is given by, n5 = 5n1

n5 = 577.694 = 388.47 Hz

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genius_generous answered 7 months ago
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