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Q.

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
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