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PART - 2 CH - 13 Oscillations — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 13 Oscillations3 Marks
Question
Show that the particle performs harmonic motion in a parabolic potential well.
✓

Answer

When a body performs simple harmonic motion then it means potential energy at a distance $x$ from the position is P.E. $= U =\frac{1}{2} k x^2$. When we draw graph of potential energy U and displacement X, we get a parabola. This type of potential function is called parabolic potential well because its shape is like a well. Simple harmonic motion is found in this type of potential well. Potential energy $U =\frac{1}{2} k x^2$
Force acting on the particle
$\begin{array}{l}F=-\frac{d}{d x}(U) \\F=-\frac{d}{d x}\left(\frac{1}{2} k x^2\right)=-\frac{1}{2} k \cdot 2 x \\F=-k x\end{array}$
It is clear that the restoring force is proportional to the displacement and its direction is towards the mean potential.
Therefore the body will perform simple harmonic motion.
Acceleration of the particle
$\begin{aligned} a & =-\frac{k}{m} x=-\omega^2 x \\ \text {or}\quad \frac{d^2 x}{d t^2} & =-\omega^2 x\end{aligned}$
and its solution will be $x= A \sin (\omega t+\phi)$. So if any potential function of the object is $U =\frac{1}{2} k x^2$ then the object will perform simple harmonic motion.

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