Question

A body of mass '$m$' performs linear S.H.M. given by the equation $x = P \sin \omega t + Q \sin \left(\omega t + \frac{\pi}{2}\right)$. The total energy of the particle at any instant is

• A. $\frac{1}{2} m \omega^2 PQ$
• B. $\frac{1}{2} m \omega^2 P^2 Q^2$
• C. $\frac{1}{2} m \omega^2 (P^2 + Q^2)$
• D. $\frac{1}{2} m \omega^2 P^2 Q^2$

Hint:

Whenever a displacement equation combines a sine and a cosine function of the same frequency ($P\sin\omega t + Q\cos\omega t$), the two components are orthogonal (at $90^\circ$). You can find the squared amplitude directly using the Pythagorean theorem ($A^2 = P^2 + Q^2$), which lets you write down the energy expression instantly!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A particle's displacement is x = P sin ωt + Q sin(ωt + π/2); find its total mechanical energy E.

Step 2: Key Formula or Approach:
Use sin(ωt + π/2) = cos ωt. Resultant amplitude A = √(P² + Q²). Total energy E = ½ m ω² A².

Step 3: Detailed Explanation:
x = P sin ωt + Q cos ωt. Amplitude A = √(P² + Q²). E = ½ m ω² (P² + Q²).

Step 4: Final Answer:
E = ½ m ω² (P² + Q²), matching option (C).