MCQ
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 $8\,cm/{s^2}$. The maximum speed of the particle is ... $ cm/s$
- A$8$
- ✓$12$
- C$16$
- D$24$
✓
Answer
Correct option: B.
$12$
b
$A = {\omega ^2}y$
$A = {\omega ^2}y$
$\omega = \sqrt {A/y} = \sqrt {\frac{8}{2}} = 2\,rad/sec$
Now ${v_{\max }} = a\omega = 6 \times 2 = 12\,cm/sec$
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