Question

If the maximum velocity and maximum acceleration of a particle executing simple harmonic motion are respectively \( 5 \, \text{m s}^{-1} \) and \( 10 \, \text{m s}^{-2} \), then the time period of oscillation of the particle is

• A. \( \pi \, \text{s} \)
• B. \( 2\pi \, \text{s} \)
• C. \( 2 \, \text{s} \)
  % Image seems to cut off option 3, assuming it was 2s
• D. \( 1 \, \text{s} \)

Hint:

For SHM: - Displacement: \( x = A\sin(\omega t + \phi) \) - Velocity: \( v = A\omega\cos(\omega t + \phi) \implies v_{max} = A\omega \) (at mean position) - Acceleration: \( a = -A\omega^2\sin(\omega t + \phi) \implies a_{max} = A\omega^2 \) (at extreme positions) - Angular frequency \( \omega \), Time period \( T = 2\pi/\omega \), Frequency \( f = 1/T = \omega/(2\pi) \). By dividing \( a_{max} \) by \( v_{max} \), we get \( \omega \).

Answer 1

The Correct Option is A

For a particle executing Simple Harmonic Motion (SHM), let A be the amplitude and \( \omega \) be the angular frequency.
Maximum velocity \( v_{max} = A\omega \).
Maximum acceleration \( a_{max} = A\omega^2 \).
Given: \( v_{max} = 5 \, \text{m s}^{-1} \implies A\omega = 5 \cdots (1) \) \( a_{max} = 10 \, \text{m s}^{-2} \implies A\omega^2 = 10 \cdots (2) \) Divide equation (2) by equation (1): \[ \frac{A\omega^2}{A\omega} = \frac{10}{5} \] \[ \omega = 2 \, \text{rad s}^{-1} \] The time period of oscillation \( T \) is related to \( \omega \) by \( T = \frac{2\pi}{\omega} \).
\[ T = \frac{2\pi}{2} = \pi \, \text{s} \] This matches option (1).