Q.A. Define linear S.H.M. Show that S.H.M. is a projection of U.C.M. on any diameter.
B. A metal sphere cools at the rate of 4°C/min when its temperature is 50°C. Find its rate of cooling at 45°C if the temperature of surroundings is 25°C. OR A. Explain analytically how stationary waves are formed. Hence show that the distance between a node and adjacent antinode is $\dfrac{\lambda}{4}$.
B. A set of 48 tuning forks is arranged in a series of descending frequencies such that each fork gives 4 beats per second with the preceding one. The frequency of the first fork is 1.5 times the frequency of the last fork. Find the frequency of the first and the 42nd tuning fork.

Project the position of a particle in uniform circular motion onto a diameter and show that this projection's motion satisfies the defining differential/algebraic equation of S.H.M.; separately apply Newton's law of cooling to scale the given cooling rate.

A. Linear S.H.M. — Definition and proof that it is the projection of U.C.M.

Definition: Linear Simple Harmonic Motion (S.H.M.) is the linear periodic motion of a body, in which the restoring force (or acceleration) is always directed towards a fixed point (the mean position) on the path and its magnitude is directly proportional to the displacement of the body from that mean position: $F=-kx$, i.e. $a=-\omega^2 x$.

Proof — S.H.M. as projection of U.C.M.:

Consider a reference particle P moving with uniform angular velocity $\omega$ on a circle of radius $A$ (called the circle of reference), centred at O. Let $M$ be the foot of the perpendicular dropped from P onto a fixed diameter, say the X-axis. As P moves around the circle, M oscillates back and forth along this diameter — M is called the projection of P.

At time $t$, let P have swept an angle $\theta=\omega t+\phi_0$ from the X-axis (where $\phi_0$ is the initial phase). The position of M, i.e. the displacement of the projection from O, is

$x=OM=OP\cos\theta=A\cos(\omega t+\phi_0)$

Differentiating twice with respect to time:

$v=\frac{dx}{dt}=-A\omega\sin(\omega t+\phi_0)$

$a=\frac{d^2x}{dt^2}=-A\omega^2\cos(\omega t+\phi_0)=-\omega^2 x$

This shows $a\propto -x$, i.e. the acceleration of M is always directed towards O and proportional to its displacement — exactly the defining condition of linear S.H.M. Hence the projection of a particle performing U.C.M., on any diameter of the circle of reference, executes linear S.H.M. with the same angular frequency $\omega$ and amplitude $A$ equal to the radius of the circle.

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