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Time period of a particle executing SHM ...

Time period of a particle executing `SHM` is `8` sec. 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

`(1)/(sqrt(2)+q)`

B

`sqrt(2)`

C

`(1)/(sqrt(2))`

D

`sqrt(2)+1`

Text Solution

Verified by Experts

The correct Answer is:
C
(c)`omega=(2pi)/(T)=(pi)/(4)s` . Therefore `y = A " sin"(pi)/(A)t` . Now put x=1 s and then x=2s.
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