← 2018 Paper 1

UPSC 2018 Maths Optional Paper 1 Q6b — Step-by-Step Solution

12 marks · Section B

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

Question

A particle moving with simple harmonic motion in a straight line has velocities v1v_1 and v2v_2 at distances x1x_1 and x2x_2 respectively from the centre of its path. Find the period of its motion.

Technique

SHM energy relation v2=ω2(A2−x2)v^2=\omega^2(A^2-x^2); subtract the two equations to cancel AA; T=2π/ωT=2\pi/\omega.

Solution

Step 1 — Velocity–displacement relation for SHM

For SHM about the centre with angular frequency ω\omega and amplitude AA, the equation of motion x¨=−ω2x\ddot x=-\omega^2x integrates to

v2=ω2 (A2−x2).v^2=\omega^2\,(A^2-x^2).

Step 2 — Apply the two data points

v12=ω2(A2−x12),v22=ω2(A2−x22).v_1^2=\omega^2(A^2-x_1^2),\qquad v_2^2=\omega^2(A^2-x_2^2).

Step 3 — Eliminate the amplitude AA

Subtract the second from the first; the A2A^2 terms cancel:

v12−v22=ω2[(A2−x12)−(A2−x22)]=ω2 (x22−x12).v_1^2-v_2^2=\omega^2\big[(A^2-x_1^2)-(A^2-x_2^2)\big]=\omega^2\,(x_2^2-x_1^2).

Hence

ω2=v12−v22x22−x12  ⇒  ω=v12−v22x22−x12.\omega^2=\frac{v_1^2-v_2^2}{x_2^2-x_1^2}\;\Rightarrow\;\omega=\sqrt{\frac{v_1^2-v_2^2}{x_2^2-x_1^2}}.

Step 4 — Period

T=2πω:T=\frac{2\pi}{\omega}:

Answer

  T=2πx22−x12v12−v22.  \boxed{\;T=2\pi\sqrt{\dfrac{x_2^2-x_1^2}{v_1^2-v_2^2}}.\;}
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