Frequency of transverse oscillations of a taut string fixed at both ends is same as that of a tuning fork and when they oscillate together, beats of frequency 5 Hz are produced. If the tension in the string is slightly decreased the beat frequency is increased to 8 Hz. If frequency of the tuning fork is 450 Hz, what is the original frequency of the string?
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a
400 Hz
b
442 Hz
c
452 Hz
d
642 Hz
answer is B.
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Detailed Solution
Frequency of transverse oscillations f∝TTherefore if f be the original frequency of the string, we can write 450−f=8⇒f=442 Hz