Q.Obtain an expression for potential energy of a particle performing S.H.M. What is the value of potential energy at

Potential energy in SHM is $\tfrac12m\omega^2x^2$ (zero at mean position, maximum at extremes); separately, sonometer-wire frequency changes are used to find the tuning fork's frequency or the wire's tension.

Option — Potential energy in SHM

For a particle of mass $m$ performing SHM with angular frequency $\omega$, displacement $x$ from the mean position, the restoring force is $F=-m\omega^2x$. The potential energy stored equals the work done against this restoring force in producing the displacement $x$:

$PE(x) = \int_0^x m\omega^2x\,dx = \frac{1}{2}m\omega^2x^2$

1. At the mean position ($x=0$): $PE = 0$ (minimum; here all the energy is kinetic).
2. At the extreme position ($x=\pm A$): $PE = \tfrac12m\omega^2A^2$ (maximum; here the particle is momentarily at rest, so all the energy is potential — this equals the total mechanical energy of the oscillator). Option — Numerical (sonometer, tuning fork frequency) A sonometer wire, originally in unison with a tuning fork (so wire frequency $f_0$ = fork frequency), has its length increased by 5%: $L' = 1.05L$. Since frequency $f \propto 1/L$ (fixed tension, mass per length): $f' = f_0\times\frac{L}{L'} = \frac{f_0}{1.05}$ This is lower than $f_0$, and the beat frequency between the (unchanged) tuning fork and the new wire frequency is 10 Hz: $f_0 - f' = f_0 - \frac{f_0}{1.05} = f_0\left(1-\frac{1}{1.05}\right) = f_0(0.047619) = 10$ $f_0 = \frac{10}{0.047619} \approx 210\ \text{Hz}$ (Check: $f' = 210/1.05 = 200\ \text{Hz}$, and $210-200=10$ beats/s ✓.) — OR (alternative) — From the SHM differential equation $\dfrac{d^2x}{dt^2}+\omega^2x=0$, with solution $x = A\sin(\omega t+\phi)$: Displacement: $x = A\sin(\omega t+\phi)$ Velocity: $v = \dfrac{dx}{dt} = A\omega\cos(\omega t+\phi) = \omega\sqrt{A^2-x^2}$ Acceleration: $a = \dfrac{dv}{dt} = -A\omega^2\sin(\omega t+\phi) = -\omega^2x$ …