Question

The period of a particle in SHM is $8\text{s}$. At $t=0$, it is at the mean position. The ratio of the distance travelled by it in the $1^{\text{st}}$ and the $2^{\text{nd}}$ second is

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

Answer 1

The correct option is D $\sqrt{2}+1$

Given that,
Time period of the particle $\left(T\right)=8\text{s}$
As the particle starts from mean position, upto $t=\frac{T}{4}=2\text{s},$ motion of particle is in same direction, hence distance is same as displacement.

When $t=1\text{sec},y_1=A\sin \left(\frac{2\pi }{T}{t}_1\right)$
$y_1=A\sin \left[\frac{2\pi }{8}\right(1\left)\right]=\frac{A}{\sqrt{2}}\dots \dots \left(1\right)$
When $t=2\text{sec}$
$y_2=A\sin \left[\frac{2\pi }{8}\right(2\left)\right]=A\dots \dots \left(2\right)$
$\frac{d_{\text{1st}}}{d_{\text{2nd}}}=\frac{y_1}{y_2-y_1}=\frac{\left(\frac{A}{\sqrt{2}}\right)}{A-\frac{A}{\sqrt{2}}}=\frac{1}{(\sqrt{2}-1)}\times \frac{(\sqrt{2}+1)}{(\sqrt{2}+1)}$
$\frac{d_{\text{1st}}}{d_{\text{2nd}}}=\sqrt{2}+1$
Hence, the correct option is (d).