Question

The time period of a particle executing SHM is $8sec$8 sec. At $T=0$, It Is at the mean position. Therefore, the ratio of the distance covered by the particle in the 1st second to the 2nd second is:

Answer 1

Step 1: Given parameters

the time period of a particle $T=8sec$

Step 2: what to find

The ratio of the distance covered by the particle in the 1st second to the 2nd second =?

Step 3: Solution

Displacement of a particle executing SHM at time t sec is given by,

$x=A\sin \omega t$ here $x=displacement,A=amplitude$

here $\omega =\frac{2\pi }{T}$

$x=A\sin \frac{2\mathrm{\pi }}{T}t$

$x=A\sin \frac{2\mathrm{\pi }}{8}t$

at time$t=1sec$, the displacement

$x_1=A\sin \frac{2\mathrm{\pi }}{8}\times 1$

$x_1=A\sin \frac{\mathrm{\pi }}{4}$

$x_1=\frac{A}{\sqrt{2}}$

at time t= 2 sec, the displacement

$x^{\text{'}}=A\sin \frac{2\mathrm{\pi }}{8}\times 2$

$x^{\text{'}}=A\sin \frac{\mathrm{\pi }}{2}$

$x^{\text{'}}=A$

The required ratio of the distance covered by the particle in the 1st second to the 2nd second

$\frac{x_1}{x_2}=\frac{x_1}{x^{\text{'}}-x_1}$

$\frac{x_1}{x_2}=\frac{{\displaystyle \frac{A}{\sqrt{2}}}}{A-\frac{A}{\sqrt{2}}}$

$\frac{x_1}{x_2}=\frac{{\displaystyle 1}}{\sqrt{2}-1}$

$\frac{x_1}{x_2}=\frac{{\displaystyle 1}}{\sqrt{2}-1}x\frac{\sqrt{2}+1}{\sqrt{2}+1}$

$\frac{x_1}{x_2}=\sqrt{2}+1$

Hence, The ratio of the distance covered by the particle in the 1st second to the 2nd second is $\sqrt{2}+1$.