Question

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

• A. Amplitude of motion is $4a$
• B. Time period of oscillations is $6\tau$
• C. Amplitude of motion is $3a$
• D. Time period of oscillations is $8\tau$

Answer 1

The correct option is B Time period of oscillations is $6\tau$

In simple harmonic, $x=A\cos \omega t$

Displacement in $\left(t\right)$ time $=A-A\cos \omega t$

for $t=\tau ;A[1-\cos \omega \tau ]=a....\left(1\right)$

for $t=2\tau ;A[1-\cos 2\omega \tau ]=3a....\left(2\right)$

On dividing $\left(1\right)÷\left(2\right)$
$\frac{A[1-\cos \omega t]}{A[1-\cos 2\omega \tau ]}=\frac{a}{3a}$

$\frac{1-\cos \omega t}{1-\cos 2\omega \tau }=\frac{1}{3}$

We know, $1-\cos 2\omega \tau =2{\sin }^2\omega \tau$

$\frac{1-\cos \omega \tau }{2{\sin }^2\omega \tau }=\frac{1}{3}$

Let assume. $k=\cos \omega \tau$

$\frac{1-k}{2(1-k^2)}=\frac{1}{3}$

$\Rightarrow \frac{1}{2(1+k)}=\frac{1}{3}$

$\Rightarrow 3=2+2k\Rightarrow k=\frac{1}{2}=\cos \omega \tau \Rightarrow \omega \tau =\frac{\pi }{3}$

At $A=2a,\omega \tau =\frac{\pi }{3}\Rightarrow \frac{2\pi }{T}\tau =\frac{\pi }{3}$

$\Rightarrow T=6\tau$

Hence, option $\left(D\right)$ is correct.