+3 Votes

The maximum velocity of a particle executing simple harmonic motion 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) \(18 \mathrm{v}\)

(b) \(12 v\)

(c) \(6 \mathrm{v}\)

(d) \(3 v\)

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1 Answer

+2 Votes
Best Answer
The correct option of this question will be (C).

Solution —

If $A$ is the amplitude of simple harmonic motion, then

$v=A \omega$

But $\omega=\frac{2 \pi}{T}$ where $T$ is the time period of oscillation.

$\therefore \quad v=A \frac{2 \pi}{T}$

When the amplitude is doubled and the time period decreased to $\frac{T}{3}$, the maximum velocity becomes

$v^{\prime}=2 A\left(\frac{2 \pi}{T / 3}\right)=6\left(A \frac{2 \pi}{T}\right)=6 v$

(using (i))

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