Question

If a particle executing simple harmonic motion with period \( T \) and displacement \( x = A \cos(\omega t) \), then the acceleration and velocity of the particle at the time \( t = \frac{T}{2} \) are respectively:

• A. 0, \( A \omega \)
• B. 0, \( A^2 \omega \)
• C. 0, \( A \omega^2 \)
• D. \( -A \), \( A^2 \omega \)
• E. \( -A \), \( A \omega^2 \)

Hint:

At \( t = \frac{T}{2} \), the particle in simple harmonic motion reaches the maximum displacement, where its velocity is zero, and the acceleration is directed towards the mean position.

Answer 1

Answer: (E)

The displacement of the particle in simple harmonic motion is given by: \[ x = A \cos(\omega t) \] where: - \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( t \) is the time.
To find the velocity and acceleration, we use the following relationships:
1. Velocity is the time derivative of displacement: \[ v = \frac{dx}{dt} = -A \omega \sin(\omega t) \] At \( t = \frac{T}{2} \), we know that the period \( T \) is the time taken for one complete cycle of the motion, and \( \omega T = 2\pi \). Thus: \[ \sin\left(\omega \frac{T}{2}\right) = \sin\left(\pi\right) = 0 \] So, the velocity at \( t = \frac{T}{2} \) is: \[ v = -A \omega \sin(\omega \frac{T}{2}) = 0 \] 2. Acceleration is the time derivative of velocity: \[ a = \frac{dv}{dt} = -A \omega^2 \cos(\omega t) \] At \( t = \frac{T}{2} \), we have: \[ \cos\left(\omega \frac{T}{2}\right) = \cos\left(\pi\right) = -1 \] Thus, the acceleration at \( t = \frac{T}{2} \) is: \[ a = -A \omega^2 \cos(\omega \frac{T}{2}) = A \omega^2 \] Therefore, at \( t = \frac{T}{2} \), the velocity is 0 and the acceleration is \( A \omega^2 \). Thus, the correct answer is: \[ \boxed{-A, A \omega^2} \]