← 2013 Paper 1

UPSC Maths 2013 Paper 1 Q5c — Solution

10 marks · Section B

Question

A body is performing S.H.M. in a straight line $OPQ$. Its velocity is zero at points $P$ and $Q$ whose distances from $O$ are $x$ and $y$ respectively and its velocity is $v$ at the mid-point between $P$ and $Q$. Find the time of one complete oscillation.

Technique

Standard SHM identification: zero-velocity points $\Rightarrow$ extremes; max-velocity midpoint $\Rightarrow$ centre; $v_{\max}=a\omega$.

Solution

Strategy. In SHM, the body oscillates between the two zero-velocity points (the extreme positions); the midpoint of these is the centre of oscillation, where the velocity attains its maximum. Use $v_{\text{max}}=a\omega$ to find the angular frequency $\omega$, then $T=2\pi/\omega$.

Step 1 — Identify amplitude and centre

The body has zero velocity at $P$ and $Q$ (distances $x$ and $y$ from $O$). In SHM, the velocity vanishes at the extreme positions of the oscillation. So $P$ and $Q$ are the extreme positions.

• Centre of oscillation $M$: midpoint of $PQ$, at distance $\dfrac{x+y}{2}$ from $O$ along the line.
• Amplitude $a$: distance from centre to either extreme = $\dfrac{|y-x|}{2}$.

Step 2 — Relate the midpoint velocity to $\omega$

For an SHM of amplitude $a$ and angular frequency $\omega$, the velocity at displacement $d$ from the centre is

At the centre $d=0$, this is maximised: $v_{\max}=a\omega$. The “mid-point between $P$ and $Q$” is $M$, the centre. So

Step 3 — Period

Answer