Q.State the differential equation of linear simple harmonic motion. Hence obtain the expression for acceleration, velocity and displacement of a particle performing linear S.H.M. A body cools from 80°C to 70°C in 5 minutes and to 62°C in the next 5 minutes. Calculate the temperature of the surroundings. OR What is meant by harmonics? Show that only odd harmonics are present as overtones in the case of an air column vibrating in a pipe closed at one end. The wavelengths of two sound waves in air are $\dfrac{81}{173}$ m and $\dfrac{81}{170}$ m. They produce 10 beats per second. Calculate the velocity of sound in air.

The SHM differential equation gives the standard $a$, $v$, $x$ expressions by direct integration; the cooling problem uses Newton's law of cooling in its 'average-temperature' form to eliminate the unknown constant. The OR alternative derives odd-harmonics-only for a closed pipe from its boundary conditions, and uses the beat-frequency formula to find the speed of sound.

Part 1 — Differential equation of linear SHM:

A particle performs linear SHM if the restoring force (and hence acceleration) is directly proportional to displacement from the mean position and always directed toward it:

$F = -kx \;\Rightarrow\; ma = -kx \;\Rightarrow\; \frac{d^2x}{dt^2} = -\frac{k}{m}x = -\omega^2 x, \quad \omega^2 = k/m.$

This is the defining differential equation of SHM: $\dfrac{d^2x}{dt^2}+\omega^2x=0$.

Acceleration: directly from the equation, $a = -\omega^2 x$.

Velocity: writing $a = v\dfrac{dv}{dx} = -\omega^2 x$ and integrating, $\int v\,dv = -\omega^2\int x\,dx \Rightarrow \dfrac{v^2}{2} = -\dfrac{\omega^2x^2}{2}+C$. Using $v=0$ at $x=A$ (extreme position) to fix $C=\omega^2A^2/2$:

$v = \pm\omega\sqrt{A^2-x^2}.$

Displacement: solving $\dfrac{d^2x}{dt^2}+\omega^2x=0$ (a standard second-order linear ODE) gives the general solution

$x = A\sin(\omega t + \phi_0),$

where $A$ is the amplitude and $\phi_0$ the initial phase, fixed by initial conditions.

Part 2 — Newton's law of cooling:

Using the average-temperature form of Newton's law of cooling, $\dfrac{\theta_1-\theta_2}{t} = k\left(\dfrac{\theta_1+\theta_2}{2}-\theta_0\right)$, where $\theta_0$ is the surrounding temperature:

First interval ($80\,^\circ\text{C}\to 70\,^\circ\text{C}$ in 5 min):

$\frac{80-70}{5} = k\left(\frac{80+70}{2}-\theta_0\right) \;\Rightarrow\; 2 = k(75-\theta_0). \quad (i)$

Second interval ($70\,^\circ\text{C}\to 62\,^\circ\text{C}$ in the next 5 min):

$\frac{70-62}{5} = k\left(\frac{70+62}{2}-\theta_0\right) \;\Rightarrow\; 1.6 = k(66-\theta_0). \quad (ii)$

Dividing (i) by (ii):

$\frac{2}{1.6} = \frac{75-\theta_0}{66-\theta_0} \;\Rightarrow\; 1.25(66-\theta_0) = 75-\theta_0 \;\Rightarrow\; 82.5 - 1.25\theta_0 = 75-\theta_0.$

$82.5-75 = 1.25\theta_0-\theta_0 \;\Rightarrow\; 7.5 = 0.25\theta_0 \;\Rightarrow\; \theta_0 = 30\,^\circ\text{C}.$

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