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 ?

A 2.0 kg particle undergoes SHM according to x = 15 sin ((pi t)/(4) +

1.	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 ?
Answer» SOLUTION :(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.