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pallavi pradeep bhardwaj Grade: 12
        

A particle mass 10 gm is executing shm with an amplitude of 0.5 m and periodic time of pi/s secThe max value of the force acting on the particle

7 years ago

Answers : (2)

Badiuddin askIITians.ismu Expert
147 Points
										

Dear pallavi


x=Xsinwt         where w=2 pi/T          and T=pi/5


 =.5sin 10t


w=√k/m


so k=w2m


       =102 *10


        =1000   gm/sec2


maximum force=kx


                       =1000*.5  m*gm/sec2


                        =.5 kg m/sec2





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7 years ago
Askiitians Expert Bharath-IITD
23 Points
										

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.



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Askiitians Experts


Adapa Bharath

7 years ago
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