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 \( x_2 = 2 \sqrt{7} \sin \left( 5t + \frac{\pi}{3} \right) \) 
Representing the displacements using phasors, we have: 
Magnitudes: \( \sqrt{7} \) and \( 2 \sqrt{7} \). Phase difference: \( 60^\circ \). 
The amplitude of the resultant simple harmonic motion (SHM) is \( 7 \). 
The phase angle \( \phi \) is calculated as: \( \phi = \tan^{-1} \left( \frac{2 \sqrt{7} \times \frac{\sqrt{3}}{2}}{\sqrt{7} + 2 \sqrt{7} \times \frac{1}{2}} \right) = \tan^{-1} \left( \frac{\sqrt{3}}{2} \right) = \tan^{-1} \left( \sqrt{3} \right) \). 
The resultant SHM is given by: \( X_R = 7 \sin \left( 5t + \phi \right) \). 
The acceleration of the resultant SHM is: \( a_R = 7 \times 25 \sin \left( 5t + \phi \right) \). 
The maximum acceleration is: \( a_{\text{max}} = 175 \, \text{cm/sec} = 175 \times 10^{-2} \, \text{m/sec} \).