Mechanics> oscillation...

the time period of theparticle executing simple harmonic motion is 8 sec.At T=0 it is at mean position.the ratio of the distance covered by the particle in 1st sec to 2nd sec is

jauneet singh , 14 Years ago

Grade 12

3 Answers

Badiuddin askIITians.ismu Expert

Last Activity: 14 Years ago

Dear jauneet let the equation of motion is X =Asin(2px/8) in first second X1 =Asin(p/4) in 2 sec X2 =Asin(p/2) so 2nd sec distance covered = X2 -X1 x1 /(X2 -X1) = Asin(p/4)/[Asin(p/2) -Asin(p/4)] =1/(v2 -1) Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Askiitians Experts Badiuddin

ankit singh

Last Activity: 4 Years ago

In simple harmonic motion, X = Asin (2πx8) In the 1st second, X1 = Asin (π4) = A√(2) So, in 2nd distance travelled by the particle will be, X2 = A - A√(2) X2 = A [ √(2)-1√(2) ] ... 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 .

ankit singh

Last Activity: 4 Years ago

In simple harmonic motion, X = Asin (2πx8) In the 1st second, X1 = Asin (π4) = A√(2) So, in 2nd distance travelled by the particle will be, X2 = A - A√(2) X2 = A [ √(2)-1√(2) ] ... is 8 sec. At t=0 it is at the mean position. The ratio of the distance so plese carifycovered by the particle in the 1st second to the 2nd second is : A .