Question 323: Oscillations

Question

The maximum velocity of a particle performing S.H.M. is '$$\mathrm{V}$$'. If the periodic time is made $$\left(\frac{1}{3}\right)^d$$ and the amplitude is doubled, then the new maximum velocity of the particle will be

Accepted Answer

Given $$\mathrm{T}^{\prime}=\frac{1}{3} \mathrm{~T}$$ and $$\mathrm{A}^{\prime}=2 \mathrm{~A}$$
$$	herefore \quad \omega^{\prime}=\frac{2 \pi}{T^{\prime}}=\frac{2 \pi}{\left(\frac{1}{3} Tight)}=\frac{6 \pi}{T}=3 \omega$$
$$\begin{aligned}
& 	herefore \quad 	ext { The new maximum velocity } \
& \qquad \begin{aligned}
\mathrm{V}^{\prime} & =\mathrm{A}^{\prime} \omega^{\prime} \
& =(2 \mathrm{~A}) 	imes(3 \omega) \
& =6 \mathrm{~A} \omega \
& =6 \mathrm{~V}
\end{aligned}
\end{aligned}$$

Answer

$$\frac{\mathrm{V}}{6}$$

Answer

$$\frac{3 \mathrm{~V}}{2}$$

Answer

$$3 \mathrm{~V}$$

Answer

$$6 \mathrm{~V}$$

Oscillations

The maximum velocity of a particle performing S.H.M. is '$\mathrm{V}'. If the periodic time is made \left(\frac{1}{3}\right)^d$ and the amplitude is doubled, then the new maximum velocity of the particle will be

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