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Q.

A body of mass m is hanging from a wire of length l. It is pulled by a distance  Δl  and left. It executes simple harmonic motion. If the young’s modulus of the wire is  Y, area of cross-section is A, then the frequency of oscillation is

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a

f=12πmlYA

b

f=12πYAlm

c

f=12πYAml

d

f=12πmYAl

answer is C.

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Detailed Solution

For a body under stress we know that its Young’s modulus can be written as   Y=FlAΔl. Here  FA is stress and Δll is strain. This restoring force can be also  written as F=−kΔl . Hence, we get k=FΔl=YAl. We know that for a simple harmonic  motion the frequency of oscillation is given by  f=12πkm . Substituting for the  value of  k we get  f=12πYAml.
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