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 $t=2\tau ,x=A-3a$
$\Rightarrow A-a=A\cos \omega \tau$
$A-3a=A\cos 2\omega \tau$
As $\cos 2\omega \tau =2{\cos }^2\omega \tau -1$
$\Rightarrow \frac{A-3a}{A}=2{\left(\frac{A-a}{A}\right)}^2-1$
$\frac{A-3a}{A}=\frac{2A^2+2a^2-4Aa-A^2}{A^2}$
$A^2-3aA=A^2+2a^2-4Aa$
$a^2=2aA$
$A=2a$
Now, $A-a=A\cos \omega \tau$
$\Rightarrow \cos \omega \tau =\frac{1}{2}$
$\frac{2\pi }{T}\tau =\frac{\pi }{3}$
$\Rightarrow T=6\tau$