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A 2 kg particle undergoes SHM according to x=1.5 sin(pie × t/4 +pie / 6) then (a) What is the total mechanical energy of the particle (b) what is the shortest time required for the particle to move from x=0.5 m to x= -0.75 m?

A 2 kg particle undergoes SHM according to x=1.5 sin(pie × t/4 +pie / 6) then (a) What is the total mechanical energy of the particle (b) what is the shortest time required for the particle to move from x=0.5 m to x= -0.75 m?

Grade:11

1 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
2 years ago
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