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

Step 1: Use the SHM velocity relation.
$v^2=\omega^2(A^2-x^2)$, where $A$ is amplitude and $\omega$ the angular frequency.
Step 2: First condition at $x=6$ cm.
$(16\pi)^2=\omega^2(A^2-36)$, that is $256\pi^2=\omega^2(A^2-36)$.
Step 3: Second condition at $x=8$ cm.
$(12\pi)^2=\omega^2(A^2-64)$, that is $144\pi^2=\omega^2(A^2-64)$.
Step 4: Subtract to find $\omega$.
$256\pi^2-144\pi^2=\omega^2(64-36)$, so $112\pi^2=28\omega^2$, giving $\omega^2=4\pi^2$.
Step 5: Find the amplitude.
From $256\pi^2=4\pi^2(A^2-36)$, we get $64=A^2-36$, so $A^2=100$ and $A=10$ cm.
Step 6: Maximum acceleration.
$a_{max}=\omega^2 A=4\pi^2\times10=40\pi^2$ cm/s$^2$.
\[ \boxed{40\pi^2} \]