1.

The potential energy of particle moving is S.H.M. is (1)/(2)kx^(2). If the frequency of the particle is n, the frequency of oscillation of P.E. is :

Answer»

N
2n
`(n)/(2)`
`n SQRT(2)`.

Solution :Here `x=r SIN omegat`.
and P.E. `=(1)/(2)kx^(2)=(1)/(2)m omega^(2)x""{because k=m omega^(2)}`
P.E. `=(1)/(2)m omega^(2).r^(2)sin^(2)omegat`
P.E. `=(1)/(2)m omega^(2)r^(2)[(1-cos 2omegat)/(2)]`
`:.` THEREFORE, frequency of P.E. is `2omega` i.e., TWICE the frequency of particle `(omega)`.
Hence correct choice is (b).


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