SHM question (1 Viewer)

[luigi]

Find the maximum acceleration of the particle

x = 3sin (2t + 5)

can someone do this for me?

 

[BlackJack]

isn't it from formula sin (nt + b)?
Therefore, n=2
a= - n^2 x
= -12 sin (2t+5)
max sin = 1, -1, whatever...
therefore max a = 12.

 

[wogboy]

If you want to derive it the long way (or understand fully how it works):

As a rule, maximum acceleration occurs at the same time zero velocity in SHM.

x = 3 sin (2t + 5) ............(A)

v = dx/dt = 6 cos (2t + 5) ....(B)

a = dv/dt = - 12 sin (2t + 5) .(C)

when v = 0,

6 cos (2t + 5) = 0

therefore

2t + 5 = pi/2

OR

2t + 5 = 3/2 * pi

(don't worry about how the "t" here is negative, all we're worried about is 2t + 5)

so we can sub this into (C) to find "a", the maximum accceleration

a = - 12 * sin pi/2 = - 12

OR

a = - 12 * sin(3/2 * pi) = -12 * -1 = 12

Therefore, the maximum acceleration is 12.