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.

this is an example of "committed" time.

committed time, as contracted time, takes need over essential and available time since it is seen as profitable work. it alludes to the time allotted to keep up a home and family. at the point when a man is commuting using committed time this individual may feel that the drive is an obligation to family, for example, strolling youngsters to class or driving a mate to work. contracted and committed time clients may feel that their commute could really compare to the commute of important or leisure time clients on the grounds that their commute is profitable work. in this manner, they might be more disposed to pick a mechanized method of movement.