The velocity of particle executing S.H.M. varies with displacement $(\mathrm{x})$ as $4…
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 $\pi=\frac{22}{7}$ )
22
44
66
88
Solution
$\begin{array}{ll}
& 4 V^2=50-\mathrm{x}^2 \\
\therefore & \mathrm{~V}^2=\frac{1}{4}\left(50-\mathrm{x}^2\right) \\
\therefore & \mathrm{V}=\frac{1}{2} \sqrt{\left(50-\mathrm{x}^2\right)}
\end{array}$
For the given particle velocity is given by, We know that, $\mathrm{V}=\omega\left(\mathrm{A}^2-\mathrm{x}_1^2\right)^{1 / 2}$
$\begin{aligned}
& \therefore \quad \omega=\frac{1}{2} \quad \Rightarrow T=\frac{2 \pi}{\omega}=4 \pi \\
& \therefore \quad T=4 \pi=4 \times \frac{22}{7}=\frac{88}{7} \\
& \therefore \quad x=88
\end{aligned}$