Question

A particle of mass m is executing oscillation about the origin on X-axis. Its potential energy is $V\left(x\right)=K|x{|}^3$. Where K is a positive constant. If the amplitude of oscillation is a, then its time period T is proportional to.

• A. $1/\sqrt{a}$
• B. a
• C. $\sqrt{a}$
• D. $a^{3/2}$

Answer 1

The correct option is A $1/\sqrt{a}$

$V\left(x\right)=K|x{|}^3$

$\therefore \left[K\right]=\frac{\left[V\right]}{[x{]}^3}=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[L^3\right]}=\left[M{L}^{-1}{T}^{-2}\right]$

Now time period may depends on

$T\propto \left(mass{)}^x\right(amplitude{)}^y(K{)}^z$

$\left[M^0{L}^{0}T\right]=\left[M{]}^x\right[L{]}^y[M{L}^{-1}{T}^{-2}{]}^z$

Equating powers we get,

$-2z=1,z=\frac{-1}{2}$

$y-z=0,y=\frac{-1}{2}$

as

$T\propto (amplitude{)}^y$

$T\propto (a{)}^{\frac{-1}{2}}$

$T\propto \frac{1}{\sqrt{a}}$