an object moves in simple harmonic motion with period 8 seconds and amplitude 5 cm. at time t=0 seconds, it's displacement d from rest is 0 cm , initially it moves in a negative direction

Answer 1

Simple harmonic motion looks like d = Acos(2πt/T) + B sin(2ππt/T) or Acos(ωt+δ)

In this case the x(0) = 0 and you are given an amplitude of 5 cm so the equation is d = 5sin(2πt/8) because the constant A must be 0 in the form of equation given and B must be Amplitude when sin is 1 (a max)

Because the velocity is negative, initially, we know that we have to change the sign as x must decrease as t increases from 0.

d = -5sin(2πt/8)

You might also know that v = 2πx_m/T cos(2πt/T) which shows why the negative sign is required at t=0 in order to have a negative v.

Answer 2

So April A.,

You haven't asked a question. Did you need to express that motion as a displacement vs. time function, perhaps? Consider the function "sin(t)" and modify that to match your given initial and ongoing conditions.

By modify, I mean: 1) is the function + or - signed as you set it up? 2) Does there need to be a leading coefficient to achieve the stated amplitude? 3) Does there need to be a coefficient on t to achieve the stated period (remember, sin expects 2*pi as a repeat distance, NOT 1) 4) Does there need to be an offset applied to the time term (either within the time coefficient or outside of it), so as to achieve a desired offset of the motion at t=0?

-- Cheers, --Mr. d.