Simple Harmonic Motion Ques 6
- A particle moves with simple harmonic motion in a straight line. In first $\tau \mathrm{sec}$, after starting from rest it travels a distance $a$ and in next $\tau$ sec, it travels $2 a$, in same direction, then
(2014 Main)
(a) amplitude of motion is $3 a$
(b) time period of oscillations is $8 \pi$
(c) amplitude of motion is $4 a$
(d) time period of oscillations is $6 \pi$
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Answer:
Correct Answer: 6.( d )
Solution:
- In SHM, a particle starts from rest, we have i.e.
$ x=A \cos \omega t, \text { at } t=0, x=A $
When $t=\tau$, then $x=A-a$
When $\quad t=2 \tau$, then
$ x=A-3 a $
On comparing Eqs. (i) and (ii), we get
$ \begin{gathered} A-a=A \cos \omega \tau \\ A-3 a=A \cos 2 \omega \tau \end{gathered} $
As $\cos 2 \omega \tau=2 \cos ^2 \omega \tau-1$
$\Rightarrow \frac{A-3 a}{A}=2\left(\frac{A-a}{A}\right)^2-1 $
$\Rightarrow \frac{A-3 a}{A}= \frac{2 A^2+2 a^2-4 A a-A^2}{A^2} $
$ A^2-3 a A =A^2+2 a^2-4 A a $
$a^2 =2 a A $
$\Rightarrow A =2 a $
$\text { Now, } A-a =A \cos \omega \tau $
$ \cos \omega \tau =1 / 2$
$\begin{aligned} \Rightarrow & \frac{2 \pi}{T} \tau & =\frac{\pi}{3} \\ \Rightarrow & T & =6 \pi\end{aligned}$
BETA
