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

The problem involves a simple harmonic oscillator with an angular frequency of 2 rad/s, subjected to an external force $$ F = \sin t $$ N. The oscillator starts at rest at its equilibrium position when $$ t = 0 $$. We need to find its position at later times, which should be proportional to one of the given options.

The equation of motion for a forced harmonic oscillator is given by Newton's second law. The general form is $$ m \frac{d^2x}{dt^2} + kx = F_{\text{ext}} $$, where $$ m $$ is the mass and $$ k $$ is the spring constant. The natural angular frequency $$ \omega $$ is related to $$ k $$ and $$ m $$ by $$ \omega^2 = \frac{k}{m} $$. Given $$ \omega = 2 $$ rad/s, we have $$ \omega^2 = 4 $$, so $$ \frac{k}{m} = 4 $$. Substituting, the equation becomes:

$$ \frac{d^2x}{dt^2} + 4x = \frac{F_{\text{ext}}}{m} $$

The external force is $$ F_{\text{ext}} = \sin t $$ N, so:

$$ \frac{d^2x}{dt^2} + 4x = \frac{\sin t}{m} $$

Since $$ m $$ is a constant, we can write the right-hand side as $$ \frac{1}{m} \sin t $$. The solution to this differential equation will give the position $$ x(t) $$.

The general solution is the sum of the homogeneous solution (free oscillation) and a particular solution (due to the external force). First, solve the homogeneous equation:

$$ \frac{d^2x_h}{dt^2} + 4x_h = 0 $$

The characteristic equation is $$ r^2 + 4 = 0 $$, which has roots $$ r = \pm 2i $$. Thus, the homogeneous solution is:

$$ x_h = A \cos 2t + B \sin 2t $$

where $$ A $$ and $$ B $$ are constants to be determined from initial conditions.

Next, find a particular solution $$ x_p $$ for the non-homogeneous equation. The forcing term is $$ \frac{1}{m} \sin t $$. Since the driving frequency (1 rad/s) is different from the natural frequency (2 rad/s), there is no resonance. Assume a particular solution of the form:

$$ x_p = C \cos t + D \sin t $$

where $$ C $$ and $$ D $$ are constants to be determined. Compute the second derivative:

$$ \frac{d^2x_p}{dt^2} = -C \cos t - D \sin t $$

Substitute $$ x_p $$ and its second derivative into the differential equation:

$$ (-C \cos t - D \sin t) + 4(C \cos t + D \sin t) = \frac{1}{m} \sin t $$

Simplify the left-hand side:

$$ (-C + 4C) \cos t + (-D + 4D) \sin t = 3C \cos t + 3D \sin t $$

Set this equal to the right-hand side:

$$ 3C \cos t + 3D \sin t = 0 \cdot \cos t + \frac{1}{m} \sin t $$

Equate coefficients of like trigonometric functions. For $$ \cos t $$:

$$ 3C = 0 \implies C = 0 $$

For $$ \sin t $$:

$$ 3D = \frac{1}{m} \implies D = \frac{1}{3m} $$

Thus, the particular solution is:

$$ x_p = \frac{1}{3m} \sin t $$

The general solution is the sum of $$ x_h $$ and $$ x_p $$:

$$ x(t) = A \cos 2t + B \sin 2t + \frac{1}{3m} \sin t $$

Apply the initial conditions. At $$ t = 0 $$, the oscillator is at its equilibrium position, so $$ x(0) = 0 $$:

$$ x(0) = A \cos 0 + B \sin 0 + \frac{1}{3m} \sin 0 = A \cdot 1 + B \cdot 0 + 0 = A $$

Set equal to zero:

$$ A = 0 $$

So the solution simplifies to:

$$ x(t) = B \sin 2t + \frac{1}{3m} \sin t $$

The oscillator is at rest at $$ t = 0 $$, so the initial velocity is zero: $$ \frac{dx}{dt}(0) = 0 $$. Compute the derivative:

$$ \frac{dx}{dt} = 2B \cos 2t + \frac{1}{3m} \cos t $$

Evaluate at $$ t = 0 $$:

$$ \frac{dx}{dt}(0) = 2B \cos 0 + \frac{1}{3m} \cos 0 = 2B \cdot 1 + \frac{1}{3m} \cdot 1 = 2B + \frac{1}{3m} $$

Set equal to zero:

$$ 2B + \frac{1}{3m} = 0 \implies 2B = -\frac{1}{3m} \implies B = -\frac{1}{6m} $$

Substitute $$ B $$ back into the solution:

$$ x(t) = -\frac{1}{6m} \sin 2t + \frac{1}{3m} \sin t $$

Factor out $$ \frac{1}{m} $$:

$$ x(t) = \frac{1}{m} \left( \frac{1}{3} \sin t - \frac{1}{6} \sin 2t \right) $$

Simplify the expression inside the parentheses:

$$ \frac{1}{3} \sin t - \frac{1}{6} \sin 2t = \frac{1}{3} \sin t - \frac{1}{6} (2 \sin t \cos t) = \frac{1}{3} \sin t - \frac{1}{3} \sin t \cos t $$

However, we can leave it as is for comparison. The position is proportional to the term in parentheses, ignoring the constant factor $$ \frac{1}{m} $$:

$$ \frac{1}{3} \sin t - \frac{1}{6} \sin 2t = \frac{1}{3} \left( \sin t - \frac{1}{2} \sin 2t \right) $$

Thus, $$ x(t) $$ is proportional to $$ \sin t - \frac{1}{2} \sin 2t $$.

Comparing with the options:

• Option A: $$ \sin t + \frac{1}{2} \cos 2t $$
• Option B: $$ \cos t - \frac{1}{2} \sin 2t $$
• Option C: $$ \sin t - \frac{1}{2} \sin 2t $$
• Option D: $$ \sin t + \frac{1}{2} \sin 2t $$

Option C matches the derived expression $$ \sin t - \frac{1}{2} \sin 2t $$.

Hence, the correct answer is Option C.