The acceleration of a particle at time is given by `A=-aomega^(2) sinomegat`. Its displacement at time t is

Options

A. `-a omega^(2)sinomegat`
B. `a sin omegat`
C. `a cosomegat`
D. `aomega sin((omegat)/(2))`

Answer

To find the displacement of a particle given its acceleration, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given acceleration**:
The acceleration \( A \) is given by:
\[ A = -a \omega^2 \sin(\omega t) \]

2. **Relate acceleration to velocity**:
Acceleration is the rate of change of velocity, which can be expressed as:
\[ A = \frac{dv}{dt} \]
Therefore, we can write:
\[ \frac{dv}{dt} = -a \omega^2 \sin(\omega t) \]

3. **Integrate to find velocity**:
To find the velocity \( v \), we integrate both sides with respect to \( t \):
\[ dv = -a \omega^2 \sin(\omega t) \, dt \]
Integrating both sides:
\[ v = \int -a \omega^2 \sin(\omega t) \, dt \]
The integral of \( \sin(\omega t) \) is:
\[ -\frac{1}{\omega} \cos(\omega t) \]
Thus, we have:
\[ v = -a \omega^2 \left(-\frac{1}{\omega} \cos(\omega t)\right) + C \]
Simplifying gives:
\[ v = a \omega \cos(\omega t) + C \]
where \( C \) is the constant of integration.

4. **Relate velocity to displacement**:
Velocity is the rate of change of displacement, which can be expressed as:
\[ v = \frac{ds}{dt} \]
Therefore, we can write:
\[ \frac{ds}{dt} = a \omega \cos(\omega t) + C \]

5. **Integrate to find displacement**:
To find the displacement \( s \), we integrate both sides with respect to \( t \):
\[ ds = (a \omega \cos(\omega t) + C) \, dt \]
Integrating both sides:
\[ s = \int (a \omega \cos(\omega t) + C) \, dt \]
The integral of \( \cos(\omega t) \) is:
\[ \frac{1}{\omega} \sin(\omega t) \]
Thus, we have:
\[ s = a \omega \left(\frac{1}{\omega} \sin(\omega t)\right) + Ct + D \]
Simplifying gives:
\[ s = a \sin(\omega t) + Ct + D \]
where \( D \) is another constant of integration.

6. **Final expression for displacement**:
The displacement of the particle at time \( t \) can be expressed as:
\[ s(t) = a \sin(\omega t) + Ct + D \]
If we assume the particle starts from rest and at the origin, we can set \( C = 0 \) and \( D = 0 \), leading to:
\[ s(t) = a \sin(\omega t) \]