Question

8

IP Two strings that are fixed at each end are identical, except that one is 0.530 cm longer than the other. Waves on these strings propagate with a speed of 34.2 m/s , and the fundamental frequency of the shorter string is 217 Hz . Part A What beat frequency is produced if each string is vibrating with its fundamental frequency? fbeat = nothing Hz Request Answer Part B Does the beat frequency in part (a) increase or decrease if the longer string is increased in length? Does the beat frequency in part (a) increase or decrease if the longer string is increased in length? increases decreases Request Answer Part C Repeat part (a), assuming that the longer string is 0.761 cm longer than the shorter string. fbeat = nothing Hz Request Answer Provide Feedback

Answer 1

Given that fundamental frequency of shorter string is 217 Hz

fundamental frequency is given by: f0 = V/(2*L)

L = V/(2*f0) = 34.2/(2*217)

L = length of shorter string = 0.0788 m

So L1 = length of longer string = L + 0.530 cm = 0.0788 + 0.00530

L1 = 0.0841 m

Now fundamental frequency of larger string will be:

f1 = V/(2*L1) = 34.2/(2*0.0841) = 203.33 Hz

So Beat frequency produced will be:

df = f - f1 = 217 - 203.33

Beat frequency = 13.67 Hz

Part B.

If length of longer string is increased than it fundamental frequeny will decrease,

And since since f1 is decreasing, So beat frequency (df = f - f1) will also increase

Part C.

L1 = length of longer string = L + 0.761 cm = 0.0788 + 0.00761

L1 = 0.08641 m

Now fundamental frequency of larger string will be:

f1 = V/(2*L1) = 34.2/(2*0.08641) = 197.89 Hz

So Beat frequency produced will be:

df = f - f1 = 217 - 197.89

Beat frequency = 19.11 Hz

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