The acceleration of a particle, given by relation, `a=-5omega^2 sin(omegat)`. At t=0, x=0 and `v= 5omega`. The displacement of this paticle at time t will be

Options

A. `-5 sin (omegat)`
B. `5 sin (omegat)`
C. `5 Cos (omegat)`
D. `-5 Cos (omegat)`

Answer

To solve the problem step by step, we will follow the given information and derive the displacement of the particle at time \( t \).

### Step 1: Write down the acceleration equation
The acceleration \( a \) of the particle is given by:
\[ a = -5\omega^2 \sin(\omega t) \]

### Step 2: Relate acceleration to velocity
Acceleration can also be expressed as the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Thus, we can write:
\[ \frac{dv}{dt} = -5\omega^2 \sin(\omega t) \]

### Step 3: Integrate to find velocity
We will integrate both sides with respect to \( t \). The limits for \( v \) will be from \( 5\omega \) (initial velocity at \( t = 0 \)) to \( v \) (velocity at time \( t \)), and the limits for \( t \) will be from \( 0 \) to \( t \):
\[ \int_{5\omega}^{v} dv = \int_{0}^{t} -5\omega^2 \sin(\omega t) dt \]
This gives:
\[ v - 5\omega = 5\omega^2 \int_{0}^{t} \sin(\omega t) dt \]

### Step 4: Compute the integral
The integral of \( \sin(\omega t) \) is:
\[ \int \sin(\omega t) dt = -\frac{1}{\omega} \cos(\omega t) \]
Thus, we have:
\[ \int_{0}^{t} \sin(\omega t) dt = -\frac{1}{\omega} \left[ \cos(\omega t) - \cos(0) \right] = -\frac{1}{\omega} \left[ \cos(\omega t) - 1 \right] \]
This simplifies to:
\[ \int_{0}^{t} \sin(\omega t) dt = \frac{1 - \cos(\omega t)}{\omega} \]

### Step 5: Substitute back into the velocity equation
Substituting the integral back, we have:
\[ v - 5\omega = 5\omega^2 \left( \frac{1 - \cos(\omega t)}{\omega} \right) \]
This simplifies to:
\[ v - 5\omega = 5\omega (1 - \cos(\omega t)) \]
Thus,
\[ v = 5\omega + 5\omega (1 - \cos(\omega t)) = 5\omega \cos(\omega t) \]

### Step 6: Relate velocity to displacement
Velocity is also the derivative of displacement \( x \) with respect to time:
\[ v = \frac{dx}{dt} \]
Thus, we can write:
\[ \frac{dx}{dt} = 5\omega \cos(\omega t) \]

### Step 7: Integrate to find displacement
Now we integrate both sides with respect to \( t \):
\[ \int dx = \int 5\omega \cos(\omega t) dt \]
The integral of \( \cos(\omega t) \) is:
\[ \int \cos(\omega t) dt = \frac{1}{\omega} \sin(\omega t) \]
Thus, we have:
\[ x = 5\omega \cdot \frac{1}{\omega} \sin(\omega t) + C \]
This simplifies to:
\[ x = 5 \sin(\omega t) + C \]

### Step 8: Determine the constant of integration
At \( t = 0 \), we know \( x = 0 \):
\[ 0 = 5 \sin(0) + C \Rightarrow C = 0 \]
Thus, the displacement becomes:
\[ x = 5 \sin(\omega t) \]

### Final Answer
The displacement of the particle at time \( t \) is:
\[ \boxed{5 \sin(\omega t)} \]