The period of a particle in SHM is `8s`. At `t=0` it is at the mean position. The ratio of the distances travled by it in the first and the second second is

Options

A. `3.2:1`
B. `4.2:1`
C. `2.4:1`
D. `1.6:`

Answer

C

Explanation

`t_(1)=1s, t_(2)=2s`
Now,`y_(1)=asinomegaxx1=asinomega`
`y_(2)=asin omegaxx2-asinomega=asin2omega-asinomega`
`:. (y_(2))/(y_(1))=(sin2omega-sinomega)/(sinomega)=(sin2omega)/(sinomega)-1`
`=(2sinomegacosomega)/(sinomega)-1=2cosomega-1`
`=2cos((2pi)/(T))-1=2cos((2pi)/(8))-1`
or `(y_(2))/(y_(1))=2cos((pi)/(4))-1=2xx(1)/(sqrt(2))-1=sqrt(2)-1`
`=1.414+1=2.414~~2.4:1`