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 32.2 m/s , and the fundamental frequency of the shorter string is 207 Hz .

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

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

c. Repeat part (a), assuming that the longer string is 0.731 cm longer than the shorter string.

Explanation / Answer

f=v/2L -> L=v/2f
L= 32.2/(2*207)= 0.0777m
longer= 0.0777+0.006= 0.0837m
f=v/2L = 32.2/(2* 0.0837)= 192.353 Hz (longer fundamental)
fbeat= 207-192.353= 14.6[Hz] approx

longer= 0.0837+0.00731= 0.0763m
f=v/2L = 32.2/(2*0.0763)= 211.009Hz
fbeat= 207-211.009= 4.009[Hz] approx.

Calculation may be wrong but concept is right

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