A simple harmonic oscillator of angular frequency `2 "rad" s^(-1)` is acted upon by an external force `F = sint N`. If the oscillator is at rest in its equilibrium position at `t = 0`, its position at later times is proportional to :-

Options

A. `sin t - 1/2 sin 2t`
B. `sin t + 1/2 cos 2t`
C. `sin t + 1/2 sin 2t`
D. `cos t - 1/2 sin 2t`

Explanation

To solve the problem step by step, we need to analyze the motion of the simple harmonic oscillator (SHO) under the influence of an external force.

### Step 1: Understanding the System
The simple harmonic oscillator has an angular frequency \( \omega = 2 \, \text{rad/s} \). The external force acting on it is given by \( F = \sin(t) \, \text{N} \). At \( t = 0 \), the oscillator is at rest in its equilibrium position.

### Step 2: Equation of Motion for the SHO
The displacement \( x \) of a simple harmonic oscillator can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant. Since the oscillator is at rest at \( t = 0 \), we can assume \( A = 0 \) initially.

### Step 3: Finding Acceleration
The acceleration \( a \) of the oscillator is given by:
\[ a = \frac{d^2x}{dt^2} = -\omega^2 x \]
Substituting \( \omega = 2 \):
\[ a = -4x \]

### Step 4: Including the External Force
The total force acting on the oscillator is the sum of the restoring force and the external force:
\[ F_{\text{total}} = F_{\text{restoring}} + F_{\text{external}} \]
The restoring force is \( F_{\text{restoring}} = -m \omega^2 x \) and the external force is \( F_{\text{external}} = \sin(t) \).

Thus, we can write:
\[ m a = -m \omega^2 x + \sin(t) \]
Dividing through by \( m \):
\[ a = -\omega^2 x + \frac{1}{m} \sin(t) \]

### Step 5: Substituting for Acceleration
From the previous step, we know that:
\[ a = -4x + \frac{1}{m} \sin(t) \]

### Step 6: Formulating the Differential Equation
Rearranging gives us the differential equation:
\[ \frac{d^2x}{dt^2} + 4x = \frac{1}{m} \sin(t) \]

### Step 7: Solving the Homogeneous Equation
The homogeneous part of the equation is:
\[ \frac{d^2x}{dt^2} + 4x = 0 \]
The characteristic equation is:
\[ r^2 + 4 = 0 \implies r = \pm 2i \]
The general solution for the homogeneous equation is:
\[ x_h(t) = C_1 \cos(2t) + C_2 \sin(2t) \]

### Step 8: Finding the Particular Solution
To find a particular solution \( x_p(t) \) for the non-homogeneous equation, we can use the method of undetermined coefficients. We assume a solution of the form:
\[ x_p(t) = A \sin(t) + B \cos(t) \]
Substituting \( x_p(t) \) into the differential equation will allow us to solve for \( A \) and \( B \).

### Step 9: Combining Solutions
The complete solution is:
\[ x(t) = x_h(t) + x_p(t) \]

### Step 10: Analyzing the Result
Since the problem asks for the position at later times, we focus on the terms that will dominate the behavior of the oscillator. The terms will include \( \sin(t) \) and \( \sin(2t) \) based on the external force and the response of the SHO.

### Final Answer
Thus, the position of the oscillator at later times is proportional to:
\[ x(t) \propto -\sin(t) + \frac{1}{2} \sin(2t) \]