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A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is `V(x) = k|x|^3` where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.
A. alphaortional to `(1)/(sqrt(a))`
B. independent of `a`
C. alphaortional to `sqrt(a)`
D. alphaortional to `a^(3//2)`

1 Answer

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Best answer
Correct Answer - A
`U = k|x|^3 rArr F = - (dU)/(dx)`
=`-3 k|x|^2` …(i)
Also, for`SHM`, `x = a sin omega t`
and `(d^2 x)/(dt^2) + omega^2 x = 0`
`rArr` acceleration, `a = (d^2 x)/(dt^2) = -omega^2 x`
`rArr F = ma`
=`m (d^2 x)/(dt^2)`
=` - m omega^2 x`
From Eqs (i) and (ii), we get `omega = sqrt((3 kx)/(m))`
`rArr T = (2 pi)/(omega) = 2 pi sqrt((m)/(3 k x))`
=`2 pi sqrt((m)/(3 k(a sin omega t)))`
`rArr T prop (1)/(sqrt(a))`.

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