Question

The time period of a particle in simple harmonic motion is 8 second. At t = 0 it is at the mean position. The ratio of the distances travelled by it in the first and second seconds is:

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{\sqrt{2}-1}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

Answer: (C)

$\frac{1}{\sqrt{2}-1}$

Answer 2

Time period of particle is given by $T=\frac{2\pi }{\omega }$ $(\because \,T=8s)$ So, $\frac{2\pi }{\omega }=8$ $\Rightarrow$ $\omega =\frac{2\pi }{8}=\frac{\pi }{4}$ Equation of SHM is $y=a\sin \omega t$ ?(i) ( $\because$ At mean position t = 0) Hence, $y=0$ For t = 1 s E (i) becomes $y=a\sin \frac{\pi }{4}\times 1=\frac{a}{\sqrt{2}}$ ?(ii) For t = 2 s E (i) becomes $y=a\sin \frac{\pi }{4}\times 2=a\sin \frac{\pi }{2}=a$ ?(iii) Now, the distance covered in 2 s is given by $a-\frac{a}{\sqrt{2}}=a\left( \frac{\sqrt{2}-1}{\sqrt{2}} \right)$ Again the ratio of the distance covered in first and second seconds is $=\frac{\frac{a}{\sqrt{2}}}{a\left( \frac{\sqrt{2-1}}{\sqrt{2}} \right)}=\frac{1}{(\sqrt{2}-1)}$