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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 \(\pi=\frac{22}{7}\) )

  1. A 22
  2. B 44
  3. C 66
  4. D 88
Verified Solution

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, \(\mathrm{V}=\omega\left(\mathrm{A}^2-\mathrm{x}_1^2\right)^{1 / 2}\)
\(\begin{aligned}
& \therefore \quad \omega=\frac{1}{2} \quad \Rightarrow T=\frac{2 \pi}{\omega}=4 \pi \\
& \therefore \quad T=4 \pi=4 \times \frac{22}{7}=\frac{88}{7} \\
& \therefore \quad x=88
\end{aligned}\)