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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) $3 v$

## 1 Answer

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

Solution &mdash;

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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