MHT CET · Physics · 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 \(\pi=\frac{22}{7}\) )
- A 22
- B 44
- C 66
- D 88
Answer & Solution
Correct Answer
(D) 88
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}
& 4 V^2=50-\mathrm{x}^2 \\
\therefore & \mathrm{~V}^2=\frac{1}{4}\left(50-\mathrm{x}^2\right) \\
\therefore & \mathrm{V}=\frac{1}{2} \sqrt{\left(50-\mathrm{x}^2\right)}
\end{array}\)
For the given particle velocity is given by,
We know that, \(\mathrm{V}=\omega\left(\mathrm{A}^2-\mathrm{x}_1^2\right)^{1 / 2}\)
\(\begin{aligned}
& \therefore \quad \omega=\frac{1}{2} \quad \Rightarrow T=\frac{2 \pi}{\omega}=4 \pi \\
& \therefore \quad T=4 \pi=4 \times \frac{22}{7}=\frac{88}{7} \\
& \therefore \quad x=88
\end{aligned}\)
& 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, \(\mathrm{V}=\omega\left(\mathrm{A}^2-\mathrm{x}_1^2\right)^{1 / 2}\)
\(\begin{aligned}
& \therefore \quad \omega=\frac{1}{2} \quad \Rightarrow T=\frac{2 \pi}{\omega}=4 \pi \\
& \therefore \quad T=4 \pi=4 \times \frac{22}{7}=\frac{88}{7} \\
& \therefore \quad x=88
\end{aligned}\)
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