← 2023 Paper 1

UPSC 2023 Maths Optional Paper 1 Q5d — Step-by-Step Solution

10 marks · Section B

Simple harmonic motion (free, damped, forced) · Dynamics & Statics · asked 9× in 14 yrs · Read the full method →

Question

A particle is moving under Simple Harmonic Motion of period $T$ about a centre $O$. It passes through the point $P$ with velocity $v$ along the direction $OP$ and $OP=p$. Find the time that elapses before the particle returns to the point $P$. What will be the value of $p$ when the elapsed time is $\dfrac{T}{2}$?

Technique

Standard SHM kinematics. The “time to return” question reduces to finding the next $t$ at which $x(t)$ matches the initial value, which yields the symmetric condition $\omega t+\phi=\pi-\phi$.

Solution

Setup. SHM with period $T$, angular frequency $\omega=2\pi/T$. Position $x(t)=A\sin(\omega t+\phi)$, velocity $\dot x=A\omega\cos(\omega t+\phi)$. Energy: $\dot x^2+\omega^2 x^2=\omega^2 A^2$.

At $t=0$: particle at $P$ ($x=p$) with velocity $\dot x=v$ (along $OP$, i.e., outward, so $v>0$).

Step 1 — Determine $A$ and $\phi$.

$A\sin\phi=p,\quad A\omega\cos\phi=v.$

Square and add: $A^2\sin^2\phi+\Bigl(\dfrac{v}{\omega}\Bigr)^2\cos^2\phi\cdot\omega^2/\omega^2$… let me redo. From the two equations:

Squaring and adding: $1=p^2/A^2+v^2/(A^2\omega^2)$, so $A^2=p^2+v^2/\omega^2$.

Also $\tan\phi=p\omega/v$ (assuming $\phi\in(0,\pi/2)$ for outward motion through $P$).

Step 2 — Time to return to $P$.

The particle at $P$ moves outward to amplitude $A$, then returns to $P$ moving inward. Set $x(t)=p$:

Solutions in one period: $\omega t+\phi=\phi$ (i.e., $t=0$) or $\omega t+\phi=\pi-\phi$ (the next time):

Use $\phi=\arctan(p\omega/v)$. Note $\pi/2-\arctan(x)=\arctan(1/x)$:

Step 3 — Find $p$ when $t=T/2$.

$\dfrac{T}{2}=\dfrac{T}{\pi}\arctan\Bigl(\dfrac{v}{p\omega}\Bigr)\Rightarrow\arctan\Bigl(\dfrac{v}{p\omega}\Bigr)=\dfrac{\pi}{2}\Rightarrow\dfrac{v}{p\omega}\to\infty\Rightarrow p\to 0$.

So $p=0$ — the particle is at the centre $O$ when it has velocity $v$ outward; from the centre, time to return is the full half-period $T/2$ (one trip out to amplitude and back).

Answer