MHT CET · Physics · Oscillations
The maximum velocity and maximum acceleration of a particle performing a linear S.H.M. is ' \(\alpha\) ' and ' \(\beta\) ' respectively. Then the path length of the particle is
- A \(\frac{\alpha^2}{\beta}\)
- B \(\frac{\beta \alpha^2}{2 \alpha^2}\)
- C \(\frac{2 \alpha^2}{\beta}\)
- D \(\frac{2 \beta}{\alpha^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{2 \alpha^2}{\beta}\)
Step-by-step Solution
Detailed explanation
For S.H.M.,
Maximum velocity, \(\alpha=\mathrm{A} \omega\)
\(\omega=\frac{\alpha}{\mathrm{A}}\)...(i)
Maximum acceleration, \(\beta=A \omega^2\)
\(\begin{aligned}
& \beta=A\left(\frac{\alpha}{A}\right)^2=\frac{\alpha^2}{A} \\
& \Rightarrow A=\frac{\alpha^2}{\beta}
\end{aligned}\)
\(\therefore \quad\) Path length \(=2 \mathrm{~A}=\frac{2 \alpha^2}{\beta}\)
Maximum velocity, \(\alpha=\mathrm{A} \omega\)
\(\omega=\frac{\alpha}{\mathrm{A}}\)...(i)
Maximum acceleration, \(\beta=A \omega^2\)
\(\begin{aligned}
& \beta=A\left(\frac{\alpha}{A}\right)^2=\frac{\alpha^2}{A} \\
& \Rightarrow A=\frac{\alpha^2}{\beta}
\end{aligned}\)
\(\therefore \quad\) Path length \(=2 \mathrm{~A}=\frac{2 \alpha^2}{\beta}\)
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