Askiitians Expert Bharath-IITD
Last Activity: 15 Years ago
Dear Pallavi,
We know for shm motion of the particle its displacement is given by y= y0 sin ωt Where ω = 2π/T = 2π/π = 2 and amplitude
y0 = 0.5 m
thus aceeleration on the particle is given by a= d2y/d2t = - y0 * ω2 * sin ωt
Maximum acceleration is amax = - y0 * ω2 = - 2 m/s2
Thus maximum force in magnitude = | m* amax| = 0.02 N
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
Adapa Bharath