Question

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 :

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

Answer 1

The Correct Option is D

To determine the position of a simple harmonic oscillator affected by an external force, we start by analyzing the given conditions and the characteristic equation of motion associated with forced oscillations.

The oscillator's natural angular frequency is given as \(2 \, \text{rad} \, \text{s}^{-1}\). The external force is \(F = \sin t \, \text{N}\). The position function \(x(t)\) for a driven harmonic oscillator with no damping can be expressed as:

\(m\frac{d^2x}{dt^2} + kx = F(t)\)

Here, \(k = m\omega^2\) where \(\omega = 2 \, \text{rad} \, \text{s}^{-1}\). The driving force is \(\sin t\). The solution involves both the homogeneous and particular solutions:

1. The homogeneous solution corresponds to the simple harmonic motion: \(x_h(t) = A \cos(\omega t) + B \sin(\omega t)\). Given the initial conditions (at rest at equilibrium at \(t = 0\)), \(A = 0\) and \(B = 0\).
2. The particular solution can be found using trial methods. Since \(F = \sin t\), a suitable trial solution is: \(x_p(t) = C \sin t + D \cos t\).

Substituting this into the differential equation, the solution requires matching coefficients for each trigonometric component, resulting in:

\(x_p(t) = \frac{1}{2} \sin t - \frac{1}{2} \sin 2t\)

Thus, the complete solution under the given initial conditions becomes proportional to:

\(x(t) = \sin t - \frac{1}{2} \sin 2t\), considering initial conditions demand zero homogeneous contribution.

The correct answer corresponds to the option: \(\sin t - \frac{1}{2}\sin 2 t\).