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  • 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, i

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

Option: 1

\sin \! t+\frac{1}{2}\sin 2t


Option: 2

\sin \! t+\frac{1}{2}\cos 2t


Option: 3

\cos \! t-\frac{1}{2}\sin 2t


Option: 4

\sin \! t-\frac{1}{2}\sin 2t


Answers (1)

best_answer

 From the equation of motion, we have

\begin{aligned} &F(t)=k x+m \ddot{x}......(1)\\ &\frac{F(t)}{m}=w_{0}^{2} x+\ddot{x} ......(2) \end{aligned}

The general solution of equation (2) consists of a sum of two parts,

The first part is solution let's say x=P(t) which satisfies equation (2), is called a particular solution.

The second part is the solution let's say x=S(t) which satisfies equation (2) with F(t)=0, is called a specific solution.

for x=P(t) 

\frac{\mathrm{d}^{2} \mathrm{P}(\mathrm{t})}{\mathrm{dt}^{2}}+\omega_{\circ}{ }^{2} \mathrm{P}(\mathrm{t})=\frac{\mathrm{F}(\mathrm{t})}{\mathrm{m}}

We try a solution of type \mathbf{P}(\mathbf{t})=\mathbf{A}_{1} \sin \omega t  whose frequency is the same as of forcing frequency which is equal to 1, and \omega_{0}= 2 \ rad/ s

So

\begin{array}{l} -\mathbf{A}_{1} \sin (t)+2^{2} \mathbf{A}_{1} \sin (t)=\frac{\sin t}{\mathbf{m}} \\\\ \Rightarrow \mathbf{A}_{1}=\frac{\frac{1}{\mathbf{m}}}{\mathbf{4}-1}=\frac{1}{\mathbf{3 m}} \end{array}

and the specific solution is given by 

\frac{\mathrm{d}^{2} \mathrm{~S}(\mathrm{t})}{\mathrm{dt}^{2}}+\omega_{0}^{2} \mathrm{~S}(\mathrm{t})=0

For which the solution is given as of SHM 

i.e \mathbf{S}(\mathbf{t})=\mathbf{A}_{2} \sin \left(\omega_{\circ} \mathbf{t}-\phi\right)=\mathbf{A}_{2} \sin (2 \mathbf{t}-\phi)

where A_2 and \phi are determined by initial conditions 

Now The general solution is given as

\mathbf{x}(\mathbf{t})=\mathbf{P}(\mathbf{t})+\mathbf{S}(\mathbf{t})=\frac{1}{\mathbf{3 M}} \sin (\mathbf{t})+\mathbf{A}_{2} \sin (\mathbf{2 t}-\phi)

\text { Given } \ \ x(t)=0 \text { at } t=0 \text { and } (\frac{d x}{d t})_{t=0}=0

So using x(t)=0 at t=0

\mathbf{0}=\mathbf{0}+\mathbf{A}_{2} \sin (\mathbf{0}-\phi) \Rightarrow \phi=2 \mathbf{k} \pi \text { where } \mathbf{k} \text { is an integer }

As \frac{d x}{d t}=\frac{1}{3 M} \cos (t)+A_{2} \times 2 \times \cos (2 t-2 k \pi)

Now using (\frac{d x}{d t})_{t=0}=0

we get

\begin{array}{l} \Rightarrow \frac{1}{3 \mathrm{M}}+\mathrm{A}_{2} \times 2=0 \\ \\ \Rightarrow \mathrm{~A}_{2}=-\frac{1}{2}* \frac{1}{3 \mathrm{M}} \end{array}

substituting the value of  A_2 in the general solution of x(t)

\mathrm{x}(\mathrm{t})=\frac{1}{3 \mathrm{M}} \sin (\mathrm{t})-(\frac{1}{2} *\frac{1}{3 \mathrm{M}} \sin (2 \mathrm{t}-2 \mathrm{k} \pi))

taking k=0

\Rightarrow x(t)=\frac{1}{3 M}\left(\sin (t)-\frac{1}{2} \sin (2 t)\right)

Hence, the correct answer is option D.

 

 

 

 

 

 

 

Posted by

Sayak

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