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UPSC 2015 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 body moving under SHM has an amplitude $a$ and time period $T$. If the velocity is trebled when the distance from mean position is $\tfrac{2}{3}a$, the period being unaltered, find the new amplitude.

Technique

Energy-conservation identity $v^2=\omega^2(A^2-x^2)$ for SHM; apply at displacement $2a/3$ before and after the velocity change (period and hence $\omega$ unchanged).

Solution

Setup. SHM with angular frequency $\omega=2\pi/T$ (unchanged). For displacement $x$ from the mean position, the standard SHM velocity formula is

where $A$ is the amplitude.

Step 1 — Original velocity at $x=2a/3$

So $v_1=\dfrac{\omega a\sqrt 5}{3}$.

Step 2 — New velocity = $3v_1$, same point $x=2a/3$

At the instant the velocity is trebled, the displacement remains $2a/3$ (this is the imposed condition; only the velocity changes — typical “impulsive impulse” framing).

After the change, with new amplitude $A'$ and same $\omega$:

Answer