Question

The period of a particle in linear SHM is 8 s. At $t=0$ it is at the mean position. Find the ratio of distance traveled by it in 1st second and 2nd second

• A. 3.2:1
• B. 2.4:1
• C. 1.6:1
• D. 4.2:1

Answer 1

Answer: (B)

2.4:1

Since particle is starting from mean position, the equation of displacement is $x=Asin\omega t$, where A is amplitude of oscillation.
Now, $\omega =\frac{2\pi }{T}=\frac{2\pi }{8}=\frac{\pi }{4}rad/s$.
Distance travelled in 1 second= $x_1=Asin\omega t=A(sin\frac{\pi }{4}\times 1)=\frac{A}{\sqrt{2}}$
Distance travelled in 2 seconds=$x_2=Asin\omega t=A(sin\frac{\pi }{4}\times 2)=A$
Distance travelled in 2nd second=$x_2-x_1=A-\frac{A}{\sqrt{2}}=A\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$
Taking ratio, we get
$Answer=\frac{x_1}{x_2-x_1}=\frac{\frac{A}{\sqrt{2}}}{A\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)}=\sqrt{2}+1=2.4:1$
Option B is correct.