Question

A simple harmonic oscillator of angular frequency $2rad{{s}^{-1}}$is acted upon by an external force $F=\sin tN$. If the oscillator is at rest in its equilibrium position at $t=0$, its position at later times is proportional to
$\begin{align}
  & \text{A}\text{. }\sin t+\dfrac{1}{2}\cos 2t \\
 & \text{B}\text{. }\cos t-\dfrac{1}{2}\sin 2t \\
 & \text{C}\text{. }\sin t-\dfrac{1}{2}\sin 2t \\
 & \text{D}\text{. }\sin t+\dfrac{1}{2}\sin 2t \\
\end{align}$

Answer 1

Hint: From the equation of motion of SHM, assume specific and particular solutions. Substitute the values of these solutions into the equation of motion and apply the initial conditions stated in the question to find their expressions thus the position of the particle $x(t)$.
Formula used: The equation of the motion:
\[\begin{align}
  & m\ddot{x}+kx=F(t) \\
 & \Rightarrow \ddot{x} + {{\omega }_{o}}^{2}x=\dfrac{F(t)}{m} \\
\end{align}\]

Complete step-by-step answer:
For the SHM motion of an object, let the mass of the object be $m$, the displacement form the mean position be $x$ and the spring constant $k=m{{\omega }_{o}}^{2}$, where ${{\omega }_{o}}$ is the angular frequency of the motion. Then, the equation of the motion of the object is given as:
\[\begin{align}
  & m\ddot{x}+kx=F(t) \\
 & \Rightarrow \ddot{x} + {{\omega }_{o}}^{2}x=\dfrac{F(t)}{m} \\
\end{align}\]
The general solution of this equation has the sum of two parts. A particular part and a specific part.
Particular part: This part satisfies the equation represented above in a general form for the value of $x(t)=P(t)$. Where, $P(t)$ is a general solution of a SHM given by:
$P(t)={{A}_{1}}\sin \omega t$, where${{A}_{1}}$ is its amplitude and ${{A}_{1}}\sin \omega $ is the frequency. Without the loss of generality $\omega =1rad/s$ can be taken. The angular frequency of the motion ${{\omega }_{o}}$ is given as $2rad/s$ .
The particular solution gives the following expression:
\[\begin{align}
  & \dfrac{{{d}^{2}}P(t)}{d{{t}^{2}}}+{{\omega }_{\circ }}^{2}P(t)=\dfrac{F(t)}{m} \\
 & -{{A}_{1}}sin(t)+{{2}^{2}}{{A}_{1}}sin(t)=\dfrac{sint}{m} \\
 & {{A}_{1}}=\dfrac{\dfrac{1}{M}}{4-1}=\dfrac{1}{3M} \\
\end{align}\]
Specific solution: It is similar to the case of a particular solution. That is a specific function $x(t)=S(t)$ when substituted in the equation of motion, the value of the force $F(t)$ vanishes. The specific function is given by:
 $S(t)={{A}_{2}}\sin ({{\omega }_{o}}t-\Phi )={{A}_{2}}\sin (2t-\phi )$
Where, ${{A}_{2}}$the amplitude and $\phi $ the phase difference is determined by the initial condition.
Substituting this value in the equation of motion gives the expression:
\[\dfrac{{{d}^{2}}S(t)}{d{{t}^{2}}}+{{\omega }_{\circ }}^{2}S(t)=0\].
The general solution of motion is given by:
\[x(t)=P(t)+S(t)=\dfrac{1}{3M}sin(t)+{{A}_{2}}sin(2t-\phi )\]
The initial condition states that the oscillator is at rest in its equilibrium position at $t=0$
Given that,
 $\begin{align}
  & x(t)=0 \\
 & t=0 \\
 & \dfrac{dx}{dt}=0 \\
\end{align}$
Therefore,
\[\begin{align}
  & {{\dfrac{dx}{dt}}_{t=0}}=0 \\
 & 0=\dfrac{1}{3M}sin(0)+{{A}_{2}}sin(0-\phi ) \\
 & 0=sin(\phi )\Rightarrow \phi =2n\pi \\
\end{align}\]
Where, n is an integer.
Differentiating the general solution of motion with respect to t we get,
\[\dfrac{dx}{dt}=\dfrac{1}{3M}cos(t)+{{A}_{2}}\times 2\times cos(2t-2k\pi )\]
At t=0, the velocity is 0
\[\begin{align}
  & \Rightarrow \dfrac{1}{3M}+{{A}_{2}}\times 2=0 \\
 & {{A}_{2}}=-\dfrac{1}{2}\dfrac{1}{3M} \\
\end{align}\]
Substituting this value of ${{A}_{2}}$ in the general solution we get:
\[x(t)=\dfrac{1}{3M}sin(t)-\dfrac{1}{2}\dfrac{1}{3M}sin(2t-2k\pi )\]
Taking k=0, the equation becomes:
\[x(t)=\dfrac{1}{3M}\left( sin(t)-\dfrac{1}{2}sin(2t) \right)\]
Therefore, option C is the correct one.

Note: Be careful while differentiating an expression. As there might be a function of x inside another one. That is :
$\begin{align}
  & y=\sin [f(x)] \\
 & \dfrac{dy}{dx}=\cos [f(x)]\dfrac{df(x)}{dx} \\
 & ~ \\
\end{align}$
For example, in this question,
$\begin{align}
  & y=\sin t \\
 & \dfrac{dy}{dx}=\cos t \\
 & ~ \\
\end{align}$ . But,
$\begin{align}
  & y=\sin [2t-\phi ] \\
 & \dfrac{dy}{dx}=\cos [2t-\phi ]\dfrac{d(2t-\phi )}{dx}=2\cos [2t-\phi ] \\
 & ~ \\
\end{align}$ .