No. Trace out the motion of one complete period, which is the time the mass takes to return to its starting position and velocity. Hint: The initial position is y = -A; what other positions does it pass through during one period?
It'd go from -A, through its equilibrium position, to A, back through equilibrium, then to –A and repeat.
Exactly right. (Think about how long it takes for each of these position changes in terms of period.)
Yes. (Assuming you used the correct displacement.)
This is true.
Also true.
No. See my first parenthetical remark above.
Each position change takes T/4, and there are 4 of them...
Right.
Still true (but not what was asked).
Let's reread the problem:
You tell me: Starting position = ? End position = ?
I'm assuming when the problem states "new" equilibrium position that they mean the .290+0.197 position, which would be the starting and ending position, and it would take a period to go from start to finish.
The ".290+0.197 position" is the initial stretched postion, not the equilibrium position. Remember that the spring is stretched and released--so the initial position is what I called y = -A.
Ohhhh. So it's 3/4T?
Nope. Review what you wrote (about successive positions) in posts #3 and 5.
Yes, that's what they mean. (As opposed to the unstretched equilibrium position of the spring before the mass was added.)
Please tell me it's 1/4T then...
Finally....
