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. `1/2`B. `1/sqrt2`C. `sqrt2`D. `(1)/(sqrt2-1)`

Time period of a particle executing `SHM` is `8` sec. At `t=0` it is a

1.	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. `1/2`B. `1/sqrt2`C. `sqrt2`D. `(1)/(sqrt2-1)`
Answer» Correct Answer - D `T=8s`, `T/4=25` The particle moves in a straight line without change in direction so displacement and distance are same in first `2s` `t=1s`, `x_1=Asinomegat` `=Asinomega=Asin((2pi)/(T))=Asin((2pi)/(8))=A/sqrt2` `t=2s`, `x_2=A` `x_2-x_1=A(1-sinomega)=A(1-sin((2pi)/(T)))` `A(1-sin((2pi)/(8)))=A(1-(1)/(sqrt2))` `(x_1)/(x_2-x_1)=(A//sqrt2)/(A(1-1/sqrt2))=(1)/(sqrt2-1)`

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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. `1/2`B. `1/sqrt2`C. `sqrt2`D. `(1)/(sqrt2-1)`

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