wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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.


___

Open in App
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


flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Dimensional Analysis
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon