← 2014 Paper 1

UPSC 2014 Maths Optional Paper 1 Q5c — Step-by-Step Solution

10 marks · Section B

Simple harmonic motion (free, damped, forced) · Dynamics & Statics · asked 8× in 13 yrs · Read the full method →

Question

A particle is performing a simple harmonic motion (S.H.M.) of period $T$ about a centre $O$ with amplitude $a$ and it passes through a point $P$, where $OP=b$ in the direction $OP$. Prove that the time which elapses before it returns to $P$ is $\dfrac{T}{\pi}\cos^{-1}\!\left(\dfrac{b}{a}\right)$.

Technique

Standard SHM parametrisation; symmetry $\sin\phi_0=\sin(\pi-\phi_0)$; trigonometric identity $\arccos+\arcsin=\pi/2$.

Solution

Setup. SHM of amplitude $a$ and period $T$, so angular frequency $\omega=2\pi/T$. Take $O$ as origin. The particle’s position: $x(t)=a\sin(\omega t+\phi_0)$ for some phase $\phi_0$.

At the initial moment ($t=0$), the particle is at $x=b$ moving in the $+$ direction (outward from $O$), so:

• $x(0)=a\sin\phi_0=b\Rightarrow\sin\phi_0=b/a$.
• $\dot x(0)=a\omega\cos\phi_0>0\Rightarrow\cos\phi_0>0$.

So $\phi_0=\arcsin(b/a)\in(0,\pi/2)$ (taking the principal branch where cosine is positive).

Step 1 — Identify next time particle is at $x=b$

The particle reaches extreme $x=a$ at time $t_1$ with $\omega t_1+\phi_0=\pi/2$, then returns. The next time $x(t)=b$ with $\sin(\omega t+\phi_0)=b/a$ corresponds to $\omega t+\phi_0=\pi-\phi_0$ (the other angle with sine $b/a$):

Step 2 — Convert to closed form

Substitute $\omega=2\pi/T$ and $\phi_0=\arcsin(b/a)$:

Use the identity $\arccos x=\dfrac{\pi}{2}-\arcsin x$ (valid for $x\in[-1,1]$):

Therefore

Answer