First slide

Simple harmonic motion

Question

From the variation of potential energy in the direction of small oscillation of a simple pendulum, find the effective spring constant for the simple pendulum, where m is mass of the bob, l is length of the simple pendulum.

Difficult

A

$\frac{m g}{l}$
B

$\frac{m g}{2 l}$
C

$\frac{2 m g}{l}$
D

$\frac{m g}{\sqrt{2 l}}$

Solution

When a particle is executing SHM, at mean position
$F = - \frac{d U}{d r} = 0$ . Also, its potential energy is minimum at equilibrium position. So, $\frac{d^{2} U}{d r^{2}} > 0$
$\frac{d U}{d r} = - F \Rightarrow \frac{d^{2} U}{d r^{2}} = - \frac{d F}{d r}$
$At equilibrium, d F = - {|\frac{d^{2} U}{d r^{2}}|}_{r = 0} d r$
Comparing this standard SHM equation, i.e.
$At equilibrium, d F = - {|\frac{d^{2} U}{d r^{2}}|}_{r = 0} d r$
In general equilibrium happens at $r = r_{0} (say)$
$∴ k_{e f f} = {|\frac{d^{2} U}{d r^{2}}|}_{r = r_{0}}$
Also, at equilibrium $F = - \frac{d U}{d r} = 0$
Now, coming to the situation given, when the bob is given a small horizontal displacement as shown.
$\begin{array}{l} U = m g y \\ y = l (1 - \cos θ) = 2 l \sin^{2} \frac{θ}{2} \\ ∴ U = 2 m g l \sin^{2} \frac{θ}{2} \end{array}$
$As θ is small \sin \frac{θ}{2} ≃ \frac{θ}{2}$
$\begin{array}{l} So U = 2 m g l (\frac{θ}{2} \cdot \frac{θ}{2}) or U = \frac{1}{2} m g l θ^{2} \\ θ ≃ \frac{x}{l} \Rightarrow U = \frac{m g}{2 l} x^{2} \\ \frac{d U}{d x} = \frac{m g}{2 l} \cdot 2 x = \frac{m g}{l} \cdot x \\ \Rightarrow \frac{d^{2} U}{d x^{2}} = \frac{m g}{l} ∴ k_{e f f} = {|\frac{d^{2} U}{d x^{2}}|}_{x = 0} = \frac{m g}{l} \end{array}$

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