Question

IP Two strings that are fixed at each end are identical, except that one is 0.520 cm longer than the other. Waves on these strings propagate with a speed of 35.2 m/s , and the fundamental frequency of the shorter string is 222 Hz .

Part A

What beat frequency is produced if each string is vibrating with its fundamental frequency?

fbeat =

Hz

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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

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Part C

Repeat part (a), assuming that the longer string is 0.751 cm longer than the shorter string.

Answer 1

λ1 =2 L λ2 =2 L + 2∆L where ∆L = 0.56cm.

v = f1 λ1 =2 f1 L = f2 λ2 =2 f2 (L + ∆L).

From the above we have L = v /2 f1 and therefore

2f2 (v/2f1 + ∆L) = v or f2 = v /2 (v/2f1 + ∆L) = 17.6 / (17.6/222 + .00520)=148.6 Hz

Thus the beat frequency will be 212 - 148.6 =63.4 Hz.

b)

If the length of longer string increases its freq will go down further and the beat frequency will increase.

c)

f2 = 17.6 / (17.6/222 + .00751) = 202.7 Hz.

The beat freq would be 212- 193.7 = 9.20Hz.