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A 2.0 kg particle undergoes SHM accordin...

A `2.0 kg` particle undergoes SHM according to `x = 15 sin ((pi t)/(4) + (pi)/(6))` (in SI units)
(a) What is the total mechanical energy of the particle ?
(b) What is the shortest time required for the particle to move from `x = 0.5m` to `x = - 0.75m` ?

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
(a) `E = (1)/(2)mv_(max)^(2) = (1)/(2)(m) omega^(2) A^(2)`
`= (1)/(2) xx 2.0 xx ((pi)/(4))^(2) xx (1.5)^(2)`
`= 1.39 J`
(b) `0.5 = 1.5sin ((pi t_(1))/(4) + (pi)/(6))`
From here find `t`.
Then, `- 0.75 = 1.5sin((pi t_(2))/(4) + (pi)/(6))`
From here find `t_(2)`.
Now, `t_(1) ~ t_(2)` is the required time.
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