Q140 / 150 · MHT CET 2021 (22 Sep Shift 2) · Physics

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

  • A
    $\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{PQ}$
  • B
    $\frac{1}{2} \frac{\mathrm{m} \omega^2}{\mathrm{P}^2 \mathrm{Q}^2}$
  • Correct answer
    C
    $\frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{P}^2+\mathrm{Q}^2\right)$
  • D
    $\frac{1}{2} \mathrm{~m}^2 \mathrm{P}^2 \mathrm{Q}^2$

Solution

$$ \mathrm{x}=\mathrm{P} \sin \omega \mathrm{t}+\mathrm{Q} \sin \left(\omega \mathrm{t}+\frac{\pi}{2}\right) $$ It can be considered as composition of two S.H.M. of amplitudes $P$ and $Q$ having phase difference $\frac{\pi}{2}$. $\therefore$ Resultant amplitude $\mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2}$ Total energy $E=\frac{1}{2} m \omega^2 R^2$ $$ =\frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{P}^2+\mathrm{Q}^2\right) $$

Hence, the correct answer is option: 'C'.

Relevant Questions