MHT CET · Physics · Oscillations
A particle executes S.H.M. of period \(\frac{2 \pi}{\sqrt{3}} \sec\) along a straight line \(4 \mathrm{~cm}\) long. The displacement of the particle at which the velocity is numerically equal to the acceleration is
- A \(2 \mathrm{~cm}\)
- B \(1 \mathrm{~cm}\)
- C \(4 \mathrm{~cm}\)
- D \(3 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(B) \(1 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
Acceleration, \(a=\omega^2 x\)
Velocity, \(v=\omega \sqrt{A^2-x^2}\)
\(
\begin{aligned}
& \omega^2 \mathrm{x}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \\
& \omega^2 \mathrm{x}^2=\mathrm{A}^2-\mathrm{x}^2 \\
& 3 \mathrm{x}^2=\mathrm{A}^2-\mathrm{x}^2 \quad\left(\therefore \omega=\frac{2 \pi}{\mathrm{T}}=\sqrt{3}\right) \\
& 4 \mathrm{x}^2=\mathrm{A}^2 \\
& \mathrm{x}=\frac{\mathrm{A}}{2}=\frac{2}{2}=1 \mathrm{~cm}
\end{aligned}
\)
Velocity, \(v=\omega \sqrt{A^2-x^2}\)
\(
\begin{aligned}
& \omega^2 \mathrm{x}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \\
& \omega^2 \mathrm{x}^2=\mathrm{A}^2-\mathrm{x}^2 \\
& 3 \mathrm{x}^2=\mathrm{A}^2-\mathrm{x}^2 \quad\left(\therefore \omega=\frac{2 \pi}{\mathrm{T}}=\sqrt{3}\right) \\
& 4 \mathrm{x}^2=\mathrm{A}^2 \\
& \mathrm{x}=\frac{\mathrm{A}}{2}=\frac{2}{2}=1 \mathrm{~cm}
\end{aligned}
\)
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