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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` |
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Answer» Correct Answer - 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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