Question

For a particle executing simple harmonic motion, if the velocities at distances \(6\text{ cm}\) and \(8\text{ cm}\) from mean position are \(16\pi\text{ cms}^{-1}\) and \(12\pi\text{ cms}^{-1}\) respectively, then maximum acceleration is

• A. \(40\pi^2\)
• B. \(4\pi^2\)
• C. \(0.4\pi^2\)
• D. \(400\pi^2\)

Hint:

In SHM, when two different positions and velocities are given, use \[ v^2=\omega^2(A^2-x^2) \] to form simultaneous equations and solve for angular frequency and amplitude.

Answer 1

The Correct Option is A

Concept: Velocity relation in SHM: \[ v^2=\omega^2(A^2-x^2) \] Maximum acceleration \[ a_{max}=\omega^2A \]

Step 1: Use first condition At \(x=6\) \[ (16\pi)^2=\omega^2(A^2-36) \] \[ 256\pi^2=\omega^2(A^2-36) \]

Step 2: Use second condition At \(x=8\) \[ (12\pi)^2=\omega^2(A^2-64) \] \[ 144\pi^2=\omega^2(A^2-64) \]

Step 3: Subtract equations \[ 112\pi^2=\omega^2(28) \] \[ \omega^2=4\pi^2 \]

Step 4: Find amplitude Using first equation \[ 256\pi^2=4\pi^2(A^2-36) \] \[ 64=A^2-36 \] \[ A^2=100 \] \[ A=10 \]

Step 5: Maximum acceleration \[ a_{max}=\omega^2A \] \[ a_{max}=4\pi^2(10) \] \[ a_{max}=40\pi^2 \] Hence \[ \boxed{40\pi^2} \]