Question 285: Oscillations

Question

The velocity of particle executing S.H.M. varies with displacement $(\mathrm{x})$ as $4 \mathrm{~V}^2=50-\mathrm{x}^2$. The time period of oscillation is $\frac{x}{7}$ second. The value of ' $x$ ' is (Take $\pi=\frac{22}{7}$)

Accepted Answer

$$\begin{aligned}
&\begin{array}{ll} 
&	ext { For the given particle velocity is given by, }\

& 4 V^2=50-\mathrm{x}^2 \
	herefore & \mathrm{~V}^2=\frac{1}{4}\left(50-\mathrm{x}^2ight) \
	herefore & \mathrm{V}=\frac{1}{2} \sqrt{\left(50-\mathrm{x}^2ight)}
\end{array}\
&	ext { We know that, } V=\omega\left(A^2-x_1^2ight)^{1 / 2}\
&\begin{aligned}
& 	herefore \quad \omega=\frac{1}{2} \quad \Rightarrow T=\frac{2 \pi}{\omega}=4 \pi \
& 	herefore \quad T=4 \pi=4 	imes \frac{22}{7}=\frac{88}{7} \
& 	herefore \quad x=88
\end{aligned}
\end{aligned}$$

Answer

22

Answer

44

Answer

66

Answer

88

Oscillations

The velocity of particle executing S.H.M. varies with displacement (\mathrm{x}) as 4 \mathrm{~V}^2=50-\mathrm{x}^2. The time period of oscillation is \frac{x}{7} second. The value of ' x ' is (Take \pi=\frac{22}{7})

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