Question

A particle is subjected to simple harmonic motions as: $ x_1 = \sqrt{7} \sin 5t \, \text{cm} $ $ x_2 = 2 \sqrt{7} \sin \left( 5t + \frac{\pi}{3} \right) \, \text{cm} $ where $ x $ is displacement and $ t $ is time in seconds. The maximum acceleration of the particle is $ x \times 10^{-2} \, \text{m/s}^2 $. The value of $ x $ is:

• A. 175
• B. 25 \(\sqrt{7}\)
• C. \( 5 \sqrt{7} \)
• D. 125

Hint:

In problems involving the superposition of simple harmonic motions, phasor addition simplifies the calculation of resultant amplitude and phase.

Answer 1

The Correct Option is A

Given: 
\( x_1 = \sqrt{7} \sin 5t \), \quad \( x_2 = 2 \sqrt{7} \sin \left( 5t + \frac{\pi}{3} \right) \) 
From phasor, the displacement is represented as: 
\[ \sqrt{7} \quad \text{and} \quad 2 \sqrt{7} \quad \text{with angle} \, 60^\circ \] \[ \text{Amplitude of resultant SHM} = 7 \] \[ \phi = \tan^{-1} \left( \frac{2 \sqrt{7} \times \frac{\sqrt{3}}{2}}{\sqrt{7} + 2 \sqrt{7} \times \frac{1}{2}} \right) = \tan^{-1} \left( \frac{\sqrt{3}}{2} \right) = \tan^{-1} \left( \sqrt{3} \right) \] \[ X_R = 7 \sin \left( 5t + \phi \right) \] \[ a_R = 7 \times 25 \sin \left( 5t + \phi \right) \] \[ a_{\text{max}} = 175 \, \text{cm/sec} = 175 \times 10^{-2} \, \text{m/sec} \]