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Grade 12Mechanics

A particle executing shm of amplitude 4 cm and T=4sec The time taken by it to move from positive extreme position to half the amplitude is

Profile image of pallavi pradeep bhardwaj
16 Years agoGrade 12
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3 Answers

Profile image of Rohith Gandhi
16 Years ago

Dear Pallavi,

 

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Please feel free to post as many doubts on our discussion forum as you can. If you find any question
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All the best Pallavi !!!




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Rohith Gandhi P

Profile image of Yati Raj
9 Years ago
At half distance from the extreme positions, K.E = PE
1/2mv2=1/2kx2
1/2 m (A\omega cos\omega t + \phi )^{2} = 1/2 (\omega^{2}m)(A/2)^{^{2}} \Rightarrow cos^{2} \omega t = 1/4 \Rightarrow (cos 2\pi /T) t = cos\pi /3 \Rightarrow (2\pi /4) t = \pi /3 \Rightarrow t = (2/3) seconds.
 Please try to understand the answer. I  felt it as the easy way and so I am sharing it with you. as the typing of equations took me time in my computer, I made it really short. If you are going to appear for the boards, you can include a lot more steps.
Profile image of Yash Chourasiya
5 Years ago
Dear Student

Please see the solution in the attachment.
643-440_ck_0f8981cf5718556c96f61417839c086c.png

I hope this solution will help you.
Thanks & Regards
Yash Chourasiya

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