Question

Time period of a particle executing $SHM$ is $8sec$. At $t=0$ it is at the mean position. The ratio of the distance covered by the particle in the $1st$ second to the $2nd$ second is :

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

Answer 1

Answer: (A), (D)

The correct options are
A $\frac{1}{\sqrt{2}+1}$
D $\sqrt{2}+1$

$Y_1=A\sin \omega =A\sin \frac{2\pi }{8}=\frac{A}{\sqrt{2}}$

Displacement when $t=2$s is $Y_2=A\sin 2\omega =A\sin 2\times \frac{\pi }{4}=A\sin \frac{\pi }{2}=A$

Displacement covered in time $2$s$=A-\frac{A}{\sqrt{2}}$

Ratio of these two displacement when $t=1$s and $t=2$s

$\frac{Y_1}{Y_2}=\frac{\frac{A}{\sqrt{2}}}{A-\frac{A}{\sqrt{2}}}$

$=\frac{\frac{A}{\sqrt{2}}}{A(1-\frac{1}{\sqrt{2}})}$

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

$=\frac{1}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}$

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