Homework Help: SHM Amplitude Calculation

1. May 7, 2010

affirmative

My question is with part c, more specifically the calculating of the amplitude part.
[PLAIN]http://img691.imageshack.us/img691/1750/shmquestion.jpg [Broken]

The answer to the question is:

I do not understand how to arrive at this conclusion in order to calculate the amplitude; it baffles me. Any poking or prodding in the right direction (or even an outright answer) would be greatly appreciated.

Last edited by a moderator: May 4, 2017
2. May 7, 2010

Je m'appelle

Isn't the amplitude

$$A = \sqrt{c^2_1 + c^2_2}$$

Where

$$x(t) = c_1cos(wt) + c_2sin(wt)$$

3. May 7, 2010

affirmative

I have no idea unfortunately. My question is how did the examiner arrive at the answer above (it is the verbatim answer for the past paper I'm currently working on).

4. May 7, 2010

annumarch

see the equilibrium point of smaller block is .05g/k = .01 m but initially it is at .03 m thus this excess distance is its amplitude . as after each oscillation it wll come back to this point .

5. May 8, 2010

affirmative

Thank you very much, that actually kind of makes sense now!