
#1
Nov2911, 10:48 PM

P: 66

1. The problem statement, all variables and given/known data
An object undergoing SHM with a period 0.9s and amplitude of 0.32 m at t=0 the object is at 0.32 m and is instantaneously at rest. Calculate the time it takes the object to go a) from 0.32 to 0.16 m b) from 0.16 m to 0 i can do a but dont understand why when i do the calculation x is not 1/2 and is 1/2 2. Relevant equations x=Asin(wt) 3. The attempt at a solution so for first part 1/2 = 0.32 cos (2+2/9)*pi * t i get 0.15 correct answer and for part b i do the same thing because it is the same distance excpt i shift the graph so t1 is at 0.16 like this 0.16 = a cos ((wt)  pi/4) 



#2
Nov3011, 03:13 AM

P: 649

If 0.15s is the correct value for the time to reach 0.16m from 0.32m, then to reach 0 from 0.32m is the time for exactly one quarter of a period. (=0.9s/4)
The time from 0.16m to zero is then just the time for the quarter period minus your first answer. 


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