Q.Prove the law of conservation of energy for a particle performing simple harmonic motion. Hence graphically show the variation of kinetic energy and potential energy w.r.t. instantaneous displacement. Two sound notes have wavelengths $\dfrac{83}{170}$ m and $\dfrac{83}{172}$ m in the air. These notes when sounded together produce 8 beats per second. Calculate the velocity of sound in the air and frequencies of the two notes. OR Explain the formation of stationary waves by analytical method. Show the formation of stationary wave diagramatically. A mass of 1 kg is hung from a steel wire of radius 0.5 mm and length 4 m. Calculate the extension produced. What should be the area of cross-section of the wire so that elastic limit is not exceeded? Change in radius is negligible. (Given: g = 9.8 m/s²; Elastic limit of steel is $2.4\times10^{8}$ N/m²; Y for steel ($Y_{steel}$) = $20\times10^{10}$ N/m²; $\pi = 3.142$)

SHM's total mechanical energy is constant, split between KE and PE; separately, superposition of two close frequencies produces beats.

Option — Part (i): Conservation of energy in SHM

Consider a particle of mass $m$ performing linear SHM with amplitude $A$ and angular frequency $\omega$ about a mean position, with displacement $x = A\sin(\omega t+\phi)$.

Kinetic energy: velocity $v = \dfrac{dx}{dt} = A\omega\cos(\omega t+\phi)$, and since $A^2-x^2 = A^2\cos^2(\omega t+\phi)$,

$KE = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2(A^2-x^2)$

Potential energy: work done against the restoring force $F=-m\omega^2 x$ in displacing the particle to $x$ is stored as PE,

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

Total energy:

$E = KE + PE = \frac{1}{2}m\omega^2(A^2-x^2) + \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2 A^2$

Since $m$, $\omega$, $A$ are all constants of the motion, $E$ is independent of $x$ and $t$ — total mechanical energy in SHM is conserved.

Graph: Plotting KE and PE against displacement $x$ (for $-A \le x \le A$): $PE = \tfrac12 m\omega^2x^2$ is an upward parabola, zero at $x=0$ (mean position) and maximum ($=\tfrac12m\omega^2A^2$) at $x=\pm A$ (extreme positions). $KE=\tfrac12m\omega^2(A^2-x^2)$ is an inverted (downward) parabola, maximum at $x=0$ and zero at $x=\pm A$. The two curves are mirror images of each other about the horizontal line $E=\tfrac12m\omega^2A^2$, and at every $x$ their sum equals this same constant value $E$ (a horizontal straight line).

Option — Part (ii): Numerical (beats)

Two notes of wavelengths $\lambda_1 = \dfrac{83}{170}$ m and $\lambda_2 = \dfrac{83}{172}$ m produce 8 beats/s.

Frequencies: $f_1 = \dfrac{v}{\lambda_1} = \dfrac{170v}{83}$, $\ f_2 = \dfrac{v}{\lambda_2} = \dfrac{172v}{83}$ (with $f_2 > f_1$ since $\lambda_2<\lambda_1$).

Beat frequency:

$f_2 - f_1 = \frac{172v}{83} - \frac{170v}{83} = \frac{2v}{83} = 8$

$v = \frac{8\times83}{2} = 332\ \text{m/s}$

Then:

$f_1 = \frac{170\times332}{83} = \frac{170}{83}\times332 = 4\times170 = 680\ \text{Hz} \qquad (\text{since } 332/83 = 4)$

$f_2 = 4\times172 = 688\ \text{Hz}$

Check: $f_2-f_1 = 688-680 = 8$ beats/s ✓.

— OR (alternative full question) —

Part (i): Formation of stationary waves (analytical method)

Consider two identical progressive waves of the same amplitude $A$, frequency, and wavelength travelling in opposite directions along a string (e.g. an incident wave and its reflection):

$y_1 = A\sin(kx-\omega t), \qquad y_2 = A\sin(kx+\omega t)$

By the principle of superposition, the resultant displacement is

$y = y_1+y_2 = A\left[\sin(kx-\omega t)+\sin(kx+\omega t)\right] = 2A\sin(kx)\cos(\omega t)$

This is a stationary (standing) wave: the spatial part $2A\sin(kx)$ gives a fixed amplitude profile at each point $x$, which then oscillates in time as $\cos(\omega t)$ — unlike a travelling wave, the waveform does not move along $x$.

• Nodes (permanently zero displacement) occur where $\sin(kx)=0$, i.e. $kx = n\pi$, or $x = n\lambda/2$ ($n=0,1,2,\dots$).
• Antinodes (maximum amplitude $2A$) occur where $\sin(kx)=\pm1$, i.e. $x = (2n+1)\lambda/4$. …