UPSC 2024 Maths Optional Paper 1 Q6b — Step-by-Step Solution

15 marks · Section B

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

Question

A particle executes simple harmonic motion such that in two of its positions, velocities are $u$ and $v$ , and the two corresponding accelerations are $f_1$ and $f_2$ . For what value(s) of $k$ is the distance between the two positions $k(v^2-u^2)$ ? Show also that the amplitude is

\frac{1}{f_2^2-f_1^2}\,\bigl[(u^2-v^2)(u^2 f_2^2-v^2 f_1^2)\bigr]^{1/2}.

Technique

Use the SHM relations $v^2=\omega^2(A^2-x^2)$ and $a=-\omega^2 x$ at each state; eliminate $\omega^2$ between them.

Solution

SHM relations. With displacement $x$ from the centre, angular frequency $\omega$ , and amplitude $A$ :

\text{velocity}^2=\omega^2(A^2-x^2),\qquad \text{acceleration}=-\omega^2 x.

So $x=-a/\omega^2$ .

Step 1 — Find $\omega^2$ .

State 1 (velocity $u$ , acceleration $f_1$ ): $u^2=\omega^2 A^2-f_1^2/\omega^2$ . State 2 (velocity $v$ , acceleration $f_2$ ): $v^2=\omega^2 A^2-f_2^2/\omega^2$ .

Subtracting:

u^2-v^2=\frac{f_2^2-f_1^2}{\omega^2}\;\Rightarrow\;\omega^2=\frac{f_2^2-f_1^2}{u^2-v^2}.

Step 2 — Distance between the two positions.

|x_2-x_1|=\left|\frac{f_2-f_1}{\omega^2}\right|=\frac{|f_2-f_1|(u^2-v^2)}{f_2^2-f_1^2}=\frac{u^2-v^2}{f_2+f_1}.

Equating to $k(v^2-u^2)=-k(u^2-v^2)$ :

k=-\frac{1}{f_1+f_2}.

Step 3 — Amplitude.

From $u^2=\omega^2 A^2-f_1^2/\omega^2$ :

A^2=\frac{u^2}{\omega^2}+\frac{f_1^2}{\omega^4}=\frac{u^2\omega^2+f_1^2}{\omega^4}.

Computing the numerator using $\omega^2=(f_2^2-f_1^2)/(u^2-v^2)$ :

u^2\omega^2+f_1^2=\frac{u^2(f_2^2-f_1^2)+f_1^2(u^2-v^2)}{u^2-v^2}=\frac{u^2 f_2^2-v^2 f_1^2}{u^2-v^2}.

A^2=\frac{(u^2 f_2^2-v^2 f_1^2)(u^2-v^2)}{(f_2^2-f_1^2)^2}.

Answer

\boxed{\;k=-\frac{1}{f_1+f_2}.\;}

\boxed{\;A=\frac{1}{f_2^2-f_1^2}\sqrt{(u^2-v^2)(u^2 f_2^2-v^2 f_1^2)}.\;}