What is the Spring Amplitude Problem?

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The Spring Amplitude Problem involves a block on a spring that oscillates between positions, passing x=3.00 cm at t=0.685s and x=-3.00 cm at t=0.886s. The confusion arises regarding the determination of angular frequency and amplitude, as some participants mistakenly assume the amplitude is simply 3 cm. However, the correct amplitude is identified as 15 cm, indicating that the block is released from a maximum position beyond x=3. The time difference between the two positions helps in calculating the period, which is essential for finding angular frequency. Understanding the initial release point is crucial for accurately solving the problem.
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Homework Statement



A block on a spring if pulled to the right and released at t=0s. It passes x=3.00 cm at t=0.685s, and it passes x=-3.00 at t=0.886s.

a. what is the angular frequency?
b. what is the amplitude?

Homework Equations



w = 2pi / T
w= sqrt k/m


The Attempt at a Solution



I don't really need a numerical value or anything, but I'm just totally confused as to how I'm supposed to figure this out?? Wouldn't the amplitude just be 3 and the difference in time is the period? The amplitude is apparently 15 =S
 
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(max-)|-----------------|(-3)------------------|(3)---------------------|(max+)
you would be right if the block oscillated between 3 and -3 but it sounds like in your case at t=0 it is released at a max point to the right of x=3 and passes through x=3 and -3 later as shown in my crude drawing above.
 
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