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 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:
Show Hint
- 175
- 25 \(\sqrt{7}\)
- \( 5 \sqrt{7} \)
- 125
The Correct Option is A
Approach Solution - 1
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:
\[ \sqrt{7} \quad \text{and} \quad 2 \sqrt{7} \quad \text{with angle} \, 60^\circ \] \[ \text{Amplitude of resultant SHM} = 7 \] \[ \phi = \tan^{-1} \left( \frac{2 \sqrt{7} \times \frac{\sqrt{3}}{2}}{\sqrt{7} + 2 \sqrt{7} \times \frac{1}{2}} \right) = \tan^{-1} \left( \frac{\sqrt{3}}{2} \right) = \tan^{-1} \left( \sqrt{3} \right) \] \[ X_R = 7 \sin \left( 5t + \phi \right) \] \[ a_R = 7 \times 25 \sin \left( 5t + \phi \right) \] \[ a_{\text{max}} = 175 \, \text{cm/sec} = 175 \times 10^{-2} \, \text{m/sec} \]
Approach Solution -2
The problem asks for the maximum acceleration of a particle that is subjected to the superposition of two simple harmonic motions (SHMs). The resultant motion will also be an SHM, and we need to find its parameters to determine the maximum acceleration.
Concept Used:
The solution is based on the principle of superposition of SHMs and the formula for acceleration in SHM.
- Superposition of SHMs: When a particle is subjected to two SHMs of the same frequency along the same line, the resultant motion is also an SHM. If the individual motions are \( x_1 = A_1 \sin(\omega t + \phi_1) \) and \( x_2 = A_2 \sin(\omega t + \phi_2) \), the resultant displacement is \( x = A_R \sin(\omega t + \delta) \). The amplitude of this resultant SHM, \( A_R \), can be found by the vector addition of the individual amplitudes (phasor method): \[ A_R = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos(\Delta\phi)} \] where \( \Delta\phi = \phi_2 - \phi_1 \) is the phase difference between the two SHMs.
- Maximum Acceleration in SHM: For a particle executing SHM with displacement \( x = A \sin(\omega t + \delta) \), the acceleration is given by \( a = -\omega^2 x \). The magnitude of the acceleration is maximum when the displacement is maximum (i.e., \( |x| = A \)). The maximum acceleration is: \[ a_{max} = \omega^2 A_R \]
Step-by-Step Solution:
Step 1: Identify the parameters of the two individual SHMs.
The given SHM equations are:
\( x_1 = \sqrt{7} \sin(5t) \, \text{cm} \)
\( x_2 = 2\sqrt{7} \sin\left(5t + \frac{\pi}{3}\right) \, \text{cm} \)
By comparing with the standard form \( x = A \sin(\omega t + \phi) \), we get:
- Amplitude of the first SHM, \( A_1 = \sqrt{7} \, \text{cm} \)
- Amplitude of the second SHM, \( A_2 = 2\sqrt{7} \, \text{cm} \)
- Angular frequency, \( \omega = 5 \, \text{rad/s} \) (same for both)
- Phase of the first SHM, \( \phi_1 = 0 \)
- Phase of the second SHM, \( \phi_2 = \frac{\pi}{3} \, \text{rad} \)
Step 2: Calculate the resultant amplitude (\( A_R \)).
First, find the phase difference between the two SHMs:
\[ \Delta\phi = \phi_2 - \phi_1 = \frac{\pi}{3} - 0 = \frac{\pi}{3} \, \text{rad} \]Now, use the formula for the resultant amplitude:
\[ A_R^2 = A_1^2 + A_2^2 + 2A_1 A_2 \cos(\Delta\phi) \]Substitute the values:
\[ A_R^2 = (\sqrt{7})^2 + (2\sqrt{7})^2 + 2(\sqrt{7})(2\sqrt{7})\cos\left(\frac{\pi}{3}\right) \] \[ A_R^2 = 7 + (4 \times 7) + 4(7)\left(\frac{1}{2}\right) \] \[ A_R^2 = 7 + 28 + 14 = 49 \]Taking the square root to find \( A_R \):
\[ A_R = \sqrt{49} = 7 \, \text{cm} \]Step 3: Calculate the maximum acceleration (\( a_{max} \)).
First, convert the resultant amplitude \( A_R \) from cm to the SI unit, meters:
\[ A_R = 7 \, \text{cm} = 0.07 \, \text{m} \]Now, use the formula for maximum acceleration, \( a_{max} = \omega^2 A_R \):
\[ a_{max} = (5 \, \text{rad/s})^2 \times (0.07 \, \text{m}) \] \[ a_{max} = 25 \times 0.07 = 1.75 \, \text{m/s}^2 \]Final Computation & Result:
The problem states that the maximum acceleration of the particle is \( x \times 10^{-2} \, \text{m/s}^2 \). We equate this with our calculated value to find \( x \).
\[ x \times 10^{-2} = 1.75 \]Solving for \( x \):
\[ x = 1.75 \times 100 = 175 \]The value of \( x \) is 175.
Top Questions on simple harmonic motion
- Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Knowing initial position \( x_0 \), and initial momentum \( p_0 \) is enough to determine the position and momentum at any time \( t \) for a simple harmonic motion with a given angular frequency \( \omega \).
Reason (R): The amplitude and phase can be expressed in terms of \( x_0 \) and \( p_0 \).
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