A mass falls freely for 2.0 m before it hits a vertical spring. The mass sticks to the

spring, and the system begins to oscillate with a period of 2.2 seconds. If you start a

stopwatch when the mass is at the lowest point of the oscillation (and you set x = 0

at that point), where is the mass when the stopwatch reads 1.0 seconds?

So, what I did is that I found (Phi not ø) to be ∏ ( pi ) but I am not very sure.

I then used period to get ω. ω=2∏/T. Also, ω=√k/m . Therefore I could find the ratio k/m

Then I used energy conservation so : mg(h+A) = 1/2 kA ^{ 2 }

I then rewrote it as k/m = ( 2g(h+A))/ A^{2}

So I got the the amplitude. Then, I used x=Acos(wt+ø)

I got the final answer to be 3.55 meters above the lowest point. I don’t care about numbers as much as I care to get the concepts right and that I got ø right because I always have some trouble with finding it. Thank you !

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