A particle of mass m and charge Q is placed in an electric filed W which varies with time t as E = `E_(0) sin omegat`. It will undergo simple harmonic motion of amplitude.

Options

A. `(QE_(0)^(2))/(m omega^(2))`
B. `(QE_(0))/(m omega^(2))`
C. `sqrt((QE_(0))/(m omega^(2)))`
D. `(QE_(0))/(m omega)`

Answer

To solve the problem of finding the amplitude of a particle of mass \( m \) and charge \( Q \) placed in a time-varying electric field \( E(t) = E_0 \sin(\omega t) \), we can follow these steps:

### Step 1: Write the expression for the electric force
The force \( F \) acting on the charged particle due to the electric field is given by:
\[ F = Q \cdot E(t) = Q \cdot E_0 \sin(\omega t) \]

### Step 2: Relate force to mass and acceleration
According to Newton's second law, the force can also be expressed as:
\[ F = m \cdot a \]
where \( a \) is the acceleration of the particle. Therefore, we can write:
\[ m \cdot a = Q \cdot E_0 \sin(\omega t) \]

### Step 3: Substitute acceleration with the derivative of velocity
Acceleration \( a \) can be expressed as the derivative of velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} \]
Thus, we can rewrite the equation as:
\[ m \frac{dv}{dt} = Q E_0 \sin(\omega t) \]

### Step 4: Integrate to find velocity
Integrating both sides with respect to time \( t \):
\[ \int m \frac{dv}{dt} dt = \int Q E_0 \sin(\omega t) dt \]
This gives us:
\[ m v = -\frac{Q E_0}{\omega} \cos(\omega t) + C_1 \]
where \( C_1 \) is the constant of integration.

### Step 5: Relate velocity to position
We know that velocity \( v \) is also the derivative of position \( x \) with respect to time:
\[ v = \frac{dx}{dt} \]
Substituting this into our equation, we have:
\[ m \frac{dx}{dt} = -\frac{Q E_0}{\omega} \cos(\omega t) + C_1 \]

### Step 6: Integrate to find position
Integrating again with respect to time:
\[ \int m \frac{dx}{dt} dt = \int \left(-\frac{Q E_0}{\omega} \cos(\omega t) + C_1\right) dt \]
This results in:
\[ m x = -\frac{Q E_0}{\omega^2} \sin(\omega t) + C_1 t + C_2 \]
where \( C_2 \) is another constant of integration.

### Step 7: Rearrange to find the amplitude
Rearranging the equation gives:
\[ x = -\frac{Q E_0}{m \omega^2} \sin(\omega t) + \frac{C_1}{m} t + \frac{C_2}{m} \]
The term \(-\frac{Q E_0}{m \omega^2} \sin(\omega t)\) represents simple harmonic motion, and the amplitude \( A \) of this motion is:
\[ A = \frac{Q E_0}{m \omega^2} \]

### Final Answer
Thus, the amplitude of the simple harmonic motion is:
\[ A = \frac{Q E_0}{m \omega^2} \]

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