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: (D)

$\sqrt{2}+1$

In simple harmonic motion,

$X=Asin\left(\frac{2\pi x}{8}\right)$

In the 1st second,

$X_1=Asin\left(\frac{\pi }{4}\right)=\frac{A}{\sqrt{2}}$

So, in 2nd distance travelled by the particle will be,

$X_2=A-\frac{A}{\sqrt{2}}$

$X_2=A\left[\frac{\sqrt{2}-1}{\sqrt{2}}\right]$=Required ratio$=\frac{X_1}{X_2}$

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