Question

A particle executing simple harmonic motion has an amplitude of 6 cm. Its acceleration at a distance of 2 cm from the mean position is$8cm ⥂ / ⥂ s^{2}$. The maximum speed of the particle is

• A. 8 cm/s
• B. 12 cm/s
• C. 16 cm/s
• D. 24 cm/s

Answer 1

Answer: (B)

12 cm/s

Answer 2

$A = \omega^{2}y$ ⇒ $\omega = \sqrt{A/y} = \sqrt{\frac{8}{2}} = 2rad/sec$

Now $v_{\max}$