Question

A particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$ Then, its time period of vibration will be

• A. $ \frac{ \beta^2}{ \alpha}$
• B. $ \frac{ 2 \pi \beta}{ \alpha} $
• C. $ \frac{ \beta^2}{ \alpha^2}$
• D. $ \frac{ \alpha}{ \beta} $

Answer 1

The Correct Option is B

To determine the time period of a particle executing simple harmonic motion (SHM), given its maximum acceleration \( \alpha \) and maximum velocity \( \beta \), we need to understand the SHM equations and relationships.

1. For SHM, the displacement \( x(t) \) in terms of angular frequency \( \omega \) and amplitude \( A \) is given by: x = A \sin(\omega t).
2. The maximum velocity \( v_{\text{max}} \) of a particle in SHM is given by: \(\beta = \omega A\).
3. The maximum acceleration \( a_{\text{max}} \) is given by: \(\alpha = \omega^2 A\).
4. We can express the amplitude \( A \) in terms of \( \beta \) and \( \alpha \) using: A = \frac{\beta}{\omega}
5. Using the maximum acceleration expression, we have: A = \frac{\alpha}{\omega^2}
6. Equating both expressions for \( A \), we get: \frac{\beta}{\omega} = \frac{\alpha}{\omega^2},
7. Solving for \( \omega \), we find: \omega = \frac{\alpha}{\beta}.
8. The time period \( T \) of SHM is related to angular frequency by: T = \frac{2\pi}{\omega}.
9. Substituting the expression for \( \omega \), we get: T = \frac{2\pi \beta}{\alpha}.

Thus, the correct time period of the vibration is provided by the formula: \(\frac{2\pi \beta}{\alpha}\).

Therefore, the correct option is: \(\frac{2\pi \beta}{\alpha}\).