Q.Show that a linear S.H.M. is the projection of a U.C.M. along any of its diameter.

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.

B. Newton's law of cooling — rate at 45°C

By Newton's law of cooling, the rate of loss of heat (and hence the rate of fall of temperature, for a body of fixed mass and specific heat) is directly proportional to the excess of the body's temperature over its surroundings:

$\left(\frac{dT}{dt}\right)\propto (T-T_s)$

Given: at $T_1=50^\circ$C, rate$_1=4^\circ$C/min, with $T_s=25^\circ$C.

$\text{rate}_1\propto (T_1-T_s)=50-25=25$

At $T_2=45^\circ$C:

$\text{rate}_2\propto (T_2-T_s)=45-25=20$

Taking the ratio:

$\frac{\text{rate}_2}{\text{rate}_1}=\frac{20}{25}=0.8$

$\text{rate}_2=4\times 0.8=3.2\ ^\circ\text{C/min}$