Question

A particle moves with simple harmonic motion in a straight line. In first $\tau$ sec, after starting from rest it travels a distance a and in next $\tau$ sec, it travels $2a$, in same direction, then

• A. amplitude of motion is 3a
• B. time period of oscillations is $8\,\tau$
• C. amplitude of motion is $4a$
• D. time period of oscillations is $6\tau$

Answer 1

Answer: (D)

As it starts from rest, we have
$x=A \cos \omega t$ At $t=0, x=A$
when $t=\tau, x=A-a$
when $t=2 \tau, x=A-3 a$
$\Rightarrow A -a=A \cos \omega\, \tau $
$A -3 a=A \cos 2 \omega \tau$
As $\cos 2 \omega \tau=2 \cos ^{2} \omega \tau-1$
$\Rightarrow \frac{A-3 a}{A}=2\left(\frac{A-a}{A}\right)^{2}-1$
$ \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}= 2aA $
$A= 2 a$
Now, $A-a=A \cos\, \omega \,\tau$
$\Rightarrow \cos \,\omega \tau=\frac{1}{2}$
$\frac{2 \pi}{T} \tau=\frac{\pi}{3}$
$\Rightarrow T=6\, \tau$