# 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

Rohith Gandhi
24 Points
14 years ago

Dear Pallavi,

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

All the best Pallavi !!!

Regards,

Rohith Gandhi P

Yati Raj
24 Points
7 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.
Yash Chourasiya
4 years ago
Dear Student

Please see the solution in the attachment.