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Question

The frequency f of vibrations of a mass m suspended from a spring of spring constant k is given by f=Cmxky, where C is a dimensionless constant. The values of x and y are respectively:

A
12,12
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B
12,12
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C
12,12
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D
12,12
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Solution

The correct option is D 12,12
We know F=KKdim(K)=[MLT2][L1]=ML0T2
dim(M)=ML0T0 dim(f)=[M0L0T1]
f=CmxKyM0L0T1=[ML0T0]k[ML0T2]y
Comparing powers of M,L and T gives,
x+y=02y=1y=12
and x=1/2

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