IP Two strings that are fixed at each end are identical, except that one is 0.60

Question

IP Two strings that are fixed at each end are identical, except that one is 0.600 cm longer than the other. Waves on these strings propagate with a speed of 30.2 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?

Part B:

Does the beat frequency in part (a) increase or decrease if the longer string is increased in length?

Part C:

Repeat part (a), assuming that the longer string is 0.801 cm longer than the shorter string.

Explanation / Answer

Part A) We know, f=v/2L

Where, f is the frequency, V is the speed, l is the length of string

L=v/2f
L= 30.2/(2*217)= 0.06958m
longer by 0.6 cm=0.006m = 0.06958m+0.006m= 0.07558m
f=v/2L = 30.2/(2*0.07558)= 199.77Hz
Now Beat frequency= 217-199.77= 17.22 Hertz

Part B) If the length of longer string increases then its freqency and the beat frequency increases.


Part C) longer by 0.801cm = 0.00801m= 0.06958+0.00801m= 0.07759m
f=v/2L = 30.2/(2*0.07759)= 194.61 Hz

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