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Question

A particle of mass m is executing oscillation about origin on x - axis. Its potential energy is U(x)=K|X|n. If

the time period T is function of its mass, amplitude(a) and k; find the value of n for Tαa12.


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Solution

[K]=[U][X]n=[ML2T2][Ln]=[ML2nT2]

T(mass)X(amp)Y(K)Z

[M0L0T]=[M]X[L]Y[ML2nT2]Z

= [MX+ZLY+2ZnZT2Z]

So, -2Z = 1 or Z=12 ................................(i)

and, X + Z = 0 or X=Z=12 ........................(ii)

and, Y + 2Z-nZ = 0 orY=1n2[from(i) and (ii)]

So, T(mass)12(amp)1n2(K)12

Given T(amp)12

So 1n2=12 n=3


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