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.

Answer 1

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