An object is undergoing shm with period 0.900 s and amplitude 0.320 m. at t = 0 the object is at x = 0.320 m and is instantaneously at rest. calculate the time it takes the object to go (a) from x = 0.320 m to x = 0.160 m and (b) from x = 0.160 m to x = 0.

For an object at Simple Harmonic Motion (SMH), we can write the equation of motion as:
x(t) = A cos[(2 pi/P) t]

where

 P is the period = 0.9

 t is time

A is the amplitude  = 0.32
Therefore for this object we have the equation:
x(t) = 0.32 cos[(2 pi/0.9) t] 

A. Calculate t from x = 0.320 m to x = 0.160 m

We know that at t = 0 the value of x would be: x=0.32 since cos 0 = 1. So we only have to find the value of t when x =0.16m. Using the equation:

0.16 = 0.32 cos[(2 pi/0.9) t] 
0.5 = cos[(2pi/0.9) t] 
pi/3 = (2 pi/0.9)t

t = 0.15 s

 
Therefore it takes 0.15 s to go from 0.32 to 0.16m

B.  Calculate t from x = 0.160 m to x = 0
The time at which x = 0 is: 
0 = 0.32 cos[2 pi/0.90)t] 
pi/2 = (2 pi/0.90)t 
t = 0.225s 

Therefore it takes 0.225s-0.150s = 0.075s to go from 0.16 to 0 m.

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