The displacement of a particle moving in S.H.M. at any instant is given by `y = a sin omegat` . The accelreation after time `t=T/4` os

Options

A. `aomega`
B. `-aomega`
C. `aomega^(2)`
D. `-aomega^(2)`

Answer

To find the acceleration of a particle moving in Simple Harmonic Motion (SHM) at time \( t = \frac{T}{4} \), we can follow these steps:

### Step 1: Understand the displacement equation
The displacement of the particle in SHM is given by:
\[ y = a \sin(\omega t) \]
where:
- \( y \) is the displacement,
- \( a \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( t \) is the time.

### Step 2: Find the expression for acceleration
The acceleration \( a \) in SHM can be derived from the displacement equation. The acceleration is the second derivative of displacement with respect to time:
\[ a = \frac{d^2y}{dt^2} \]
First, we find the velocity \( v \) by differentiating the displacement:
\[ v = \frac{dy}{dt} = \frac{d}{dt}(a \sin(\omega t)) = a \omega \cos(\omega t) \]
Next, we differentiate the velocity to find the acceleration:
\[ a = \frac{dv}{dt} = \frac{d}{dt}(a \omega \cos(\omega t)) = -a \omega^2 \sin(\omega t) \]

### Step 3: Substitute \( t = \frac{T}{4} \)
The time period \( T \) is related to the angular frequency by \( T = \frac{2\pi}{\omega} \). Therefore, at \( t = \frac{T}{4} \):
\[ \omega t = \omega \left(\frac{T}{4}\right) = \frac{\omega T}{4} = \frac{2\pi}{4} = \frac{\pi}{2} \]
Now, substituting this into the acceleration formula:
\[ a = -a \omega^2 \sin\left(\frac{\pi}{2}\right) \]
Since \( \sin\left(\frac{\pi}{2}\right) = 1 \):
\[ a = -a \omega^2 \cdot 1 = -a \omega^2 \]

### Final Result
Thus, the acceleration of the particle at time \( t = \frac{T}{4} \) is:
\[ a = -a \omega^2 \]