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A particle oscillates simple harmonically along a straight line with period 8 seconds and amplitude `8 sqrt(2)`m. It starts from the mean position, then the ratio of the distances travelled by it in the second second and first second of its motion isA. `sqrt(2)`B. 2C. `(sqrt(2)-1)`D. `sqrt(3)` |
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Answer» Correct Answer - C Given, `8 sqrt(2)`m and T = 8 s and `omega = (2pi)/(8)=(pi)/(4)` The particle starts from the mean position. `therefore" "x = A sin omega t` `therefore` The distance travelled by the particle in one second is `therefore" "x_(1)=8 sqrt(2)sin((pi)/(4)xx1)` `= 8 sqrt(2)xx(1)/(sqrt(2))=8m = s_(1)` and the distance travelled by the particle in 2s is `x_(2)=8 sqrt(2)sin((pi)/(4)xx2)=8 sqrt(2)m` `therefore` Distance travelled in second second `(s_(2))` `=x_(2)-x_(1)=8 sqrt(2)-8 =(sqrt(2)-1)m` `therefore" "(s_(2))/(s_(1))=(8 sqrt(2)-1)/(8)=sqrt(2)-1` |
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