Question

The maximum velocity and maximum acceleration of a particle performing a linear S.H.M. are \(\alpha\) and \(\beta\) respectively. Then, path length of the particle is

• A. \(\frac{\alpha ^{2}}{\beta }\)
• B. \(\frac{\beta }{2\alpha ^{2}}\)
• C. \(\frac{2\alpha ^{2}}{\beta }\)
• D. \(\frac{2\beta }{\alpha ^{2}}\)

Hint:

Maximum velocity is Aomega and maximum acceleration is Aomega squared; the path length is 2A.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
\(v_{max}=A\omega=\alpha\) and \(a_{max}=A\omega^{2}=\beta\).

Step 2: Find the Amplitude
Divide: \(\omega=\dfrac{\beta}{\alpha}\). Then \(A=\dfrac{\alpha}{\omega}=\dfrac{\alpha^{2}}{\beta}\).

Step 3: Path Length
The particle moves from \(-A\) to \(+A\), so path length \(=2A=\dfrac{2\alpha^{2}}{\beta}\).

Final Answer:
The path length of the particle is 2alpha^2/beta. \[ \boxed{\text{(C) }\dfrac{2\alpha^{2}}{\beta}} \]