Question

Particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that is at a distance \(\frac{2A}{3}\) from the equilibrium position. The new amplitude of the motion is

• A. \(\frac{A}{3}\sqrt{41}\)
• B. \(3A\)
• C. \(\sqrt{3}A\)
• D. \(\frac{7A}{3}\)

Hint:

Use v squared = w squared (A squared minus x squared) before and after the speed changes, keeping x and w fixed.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
In SHM, \(v^{2}=\omega^{2}(A^{2}-x^{2})\). Tripling the speed at the same position changes the amplitude, while \(\omega\) stays the same.

Step 2: Original speed at x = 2A/3
\(v^{2}=\omega^{2}\left(A^{2}-\dfrac{4A^{2}}{9}\right)=\dfrac{5}{9}\omega^{2}A^{2}\)

Step 3: After tripling
\(v'^{2}=9v^{2}=5\omega^{2}A^{2}\)

Step 4: New amplitude
\(A'^{2}=x^{2}+\dfrac{v'^{2}}{\omega^{2}}=\dfrac{4A^{2}}{9}+5A^{2}=\dfrac{49A^{2}}{9}\), so \(A'=\dfrac{7A}{3}\), option (D).

Final Answer:
The new amplitude is \(\dfrac{7A}{3}\). \[ \boxed{\text{(D) }\dfrac{7A}{3}} \]