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:
We are given the displacement equation $x(t) = P \sin \omega t + Q \sin \left(\omega t + \frac{\pi}{2}\right)$ for a particle of mass $m$ undergoing Simple Harmonic Motion (S.H.M.). We need to find an expression for its total mechanical energy.

Step 2: Key Formula or Approach:
The total mechanical energy $E$ of a particle performing S.H.M. is constant at all times and is given by the formula: $$E = \frac{1}{2} m \omega^2 A^2$$ where $A$ is the resultant net amplitude of the combined harmonic wave function. We can find this amplitude by simplifying the expression using the trigonometric identity $\sin\left(\theta + \frac{\pi}{2}\right) = \cos\theta$.

Step 3: Detailed Explanation:
Let's simplify the given displacement equation: $$x = P \sin \omega t + Q \sin \left(\omega t + \frac{\pi}{2}\right)$$ Using the identity $\sin\left(\omega t + \frac{\pi}{2}\right) = \cos \omega t$: $$x = P \sin \omega t + Q \cos \omega t$$ This expression represents the superposition of two perpendicular harmonic oscillations with a phase difference of exactly $90^\circ$ ($\frac{\pi}{2}$). The resultant net amplitude $A$ for two out-of-phase component vectors is given by the vector addition formula: $$A = \sqrt{P^2 + Q^2 + 2PQ\cos\left(\frac{\pi}{2}\right)}$$ Since $\cos\left(\frac{\pi}{2}\right) = 0$, the expression simplifies to: $$A = \sqrt{P^2 + Q^2} \implies A^2 = P^2 + Q^2$$ Now, substitute this definition of $A^2$ into our total energy equation: $$E = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} m \omega^2 (P^2 + Q^2)$$ This matches option (C).

Step 4: Final Answer:
The total energy of the particle is $\frac{1}{2} m \omega^2 (P^2 + Q^2)$, which corresponds to option (C).