Q. The maximum velocity of a particle executing SHM is $V.$ If the amplitude is doubled and the time period of oscillation decreased to $1/3$ of its original value, the maximum velocity becomes

A) $18\ V$
B) $12\ V$
C) $6\ V$
D) $3\ V$
Explanation

(c) Maximum velocity $V_{\max} = a\omega$
$\omega = \frac{2\pi}{T}\ \therefore V_{\max} = \frac{2\pi a}{T},\ i.e.,\ V \propto \frac{a}{T}$
$\therefore\frac{V_{1}}{V_{2}} = \frac{a_{1}}{a_{2}} \times \frac{T_{2}}{T_{1}} = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$
$\therefore V_{2} = 6V_{1} = 6V$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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