The position x of particle moving along x-axis varies with time t as `x=Asin(omegat)` where A and `omega` are positive constants. The acceleration a of particle varies with its position (x) as

Options

A. `a = Ax`
B. `a = - omega^(2)x`
C. `a = A omegax`
D. `a = omega^(2)x A`

Answer

To solve the problem, we need to find the expression for the acceleration \( a \) of a particle whose position \( x \) varies with time \( t \) as given by the equation:

\[ x = A \sin(\omega t) \]

where \( A \) and \( \omega \) are positive constants.

### Step 1: Find the Velocity

The velocity \( v \) of the particle is the rate of change of position with respect to time. We can find it by differentiating \( x \) with respect to \( t \):

\[ v = \frac{dx}{dt} \]

Using the chain rule, we differentiate \( x = A \sin(\omega t) \):

\[ v = A \frac{d}{dt}(\sin(\omega t)) = A \cos(\omega t) \cdot \frac{d}{dt}(\omega t) = A \omega \cos(\omega t) \]

### Step 2: Find the Acceleration

The acceleration \( a \) of the particle is the rate of change of velocity with respect to time. We can find it by differentiating \( v \) with respect to \( t \):

\[ a = \frac{dv}{dt} \]

Differentiating \( v = A \omega \cos(\omega t) \):

\[ a = A \omega \frac{d}{dt}(\cos(\omega t)) = A \omega (-\sin(\omega t)) \cdot \frac{d}{dt}(\omega t) = -A \omega^2 \sin(\omega t) \]

### Step 3: Express Acceleration in Terms of Position \( x \)

From the original position equation \( x = A \sin(\omega t) \), we can express \( \sin(\omega t) \) in terms of \( x \):

\[ \sin(\omega t) = \frac{x}{A} \]

Now, substituting this back into the acceleration equation:

\[ a = -A \omega^2 \left(\frac{x}{A}\right) \]

This simplifies to:

\[ a = -\omega^2 x \]

### Final Result

Thus, the acceleration \( a \) of the particle varies with its position \( x \) as:

\[ a = -\omega^2 x \]