Question

The time period of a particle executing S.H.M. is $8 \, s$ . At $t=0$ it is at the mean position. The ratio of the distance covered by the particle in the $1^{s t}$ second to the $2^{n d}$ second is

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

Answer 1

Answer: (D)

$T=8s$
$\omega =\frac{2 \pi }{T}=\frac{\pi }{4}$
Equation of S.H.M. $y=Asin \frac{\pi }{4}t$
$y\left(1\right)=Asin \left(\frac{\pi }{4}\right)=\frac{A}{\sqrt{2}}$
$y\left(2\right)=Asin \left(\frac{\pi }{2}\right)=A$
Distance travelled in $2^{n d}$ second $=y\left(2\right)-y\left(1\right)$
$=A-\frac{A}{\sqrt{2}}$
So the ratio of distance travelled in second to the first second would be $=\frac{A - \frac{A}{\sqrt{2}}}{\frac{A}{\sqrt{2}}}$
$=\frac{\sqrt{q} - 1}{1}=\frac{\left(\sqrt{2} - 1\right) \, \left(\sqrt{2} + 1\right)}{\sqrt{2} + 1}=\left(\frac{1}{\sqrt{2} + 1}\right)$
So the ratio of the distance travelled in the $1^{s t}$ to $2^{n d}$ second $=\sqrt{2}+1$