Q. A particle of mass $ \, m$ is executing oscillations about the origin on the x-axis. Its potential energy is $U\left(x\right)=k\left(\left[x\right]\right)^{3}$ , where $k$ is a positive constant. If the amplitude of oscillation is $a$ , then its time period $T \, $ is
NTA AbhyasNTA Abhyas 2022
Solution:
Given,
$U=k\left|x\right|^{3}$
From conservative force and potential energy relation, we know,
$\Rightarrow F=-\frac{d U}{d x}=-3k\left(\left|x\right|\right)^{2}...\left(1\right)$
Also, for SHM $x=asin\omega t$ and $\frac{d^{2} x}{d t^{2}}+\omega ^{2}x=0$
$\Rightarrow $ Acceleration $a=\frac{d^{2} x}{d t^{2}}=-\omega ^{2}x$
$\Rightarrow F=ma$
$\Rightarrow F=-m\left(\omega \right)^{2}x...\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$ we get $\omega =\sqrt{\frac{3 k x}{m}}$
$\Rightarrow T=\frac{2 \pi }{\omega }=2\pi \sqrt{\frac{m}{3 k x}}=2\pi \sqrt{\frac{m}{3 k \left(a s i n \omega t\right)}}$ $\Rightarrow T \propto \frac{1}{\sqrt{a}}$ .
$U=k\left|x\right|^{3}$
From conservative force and potential energy relation, we know,
$\Rightarrow F=-\frac{d U}{d x}=-3k\left(\left|x\right|\right)^{2}...\left(1\right)$
Also, for SHM $x=asin\omega t$ and $\frac{d^{2} x}{d t^{2}}+\omega ^{2}x=0$
$\Rightarrow $ Acceleration $a=\frac{d^{2} x}{d t^{2}}=-\omega ^{2}x$
$\Rightarrow F=ma$
$\Rightarrow F=-m\left(\omega \right)^{2}x...\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$ we get $\omega =\sqrt{\frac{3 k x}{m}}$
$\Rightarrow T=\frac{2 \pi }{\omega }=2\pi \sqrt{\frac{m}{3 k x}}=2\pi \sqrt{\frac{m}{3 k \left(a s i n \omega t\right)}}$ $\Rightarrow T \propto \frac{1}{\sqrt{a}}$ .