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Two identical strings, of identical lengths of 2.00 m and linear mass density of H 0.0065 kg/m,are fixed on both ends. String A is under a tension of 120.00 N. String B is under a tension of 130.00 N. They are each plucked and produce sound at the n-10 mode. What is the beat frequency? Select the correct answer O 10 Hz Your Answer Hz O 20 Hz O 5 Hz O 15 Hz

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Answer #1

answer)

we know the formula

f=nv/2l

n=10

l=2m

v=\sqrt{}T/u

so for f1

v=\sqrt{}120/.0065=136m/s

so using eqn 1

f1=10*1360/4=340Hz

for f2

v=\sqrt{}130/0.0065=141Hz

f2=10*141.42/4=354Hz

beat frequncy=f2-f1=354-340=14Hz

so the correct option is 15Hz. ( since its closer 14Hz)

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