- #71
- 42,149
- 10,208
Remember that a is ##\ddot x##. The standard form of the SHM ODE is ##\ddot x+\omega^2x=0##, but that is taking x as offset from the equilibrium position. Because your x is not measured from there the form you have is ##\ddot x+\omega^2x=C##.songoku said:I get:
$$T=mg \sin \alpha + ma \cos \alpha - ma$$
and
$$a=\frac{mg \sin \alpha -kx}{M + 2m(1-\cos \alpha)}$$
How to proceed to find period? Thanks
You could use the equilibrium position you found to adjust your x to be an offset from there. If you have done everything correctly c should disappear. Or you could just trust that it will disappear and ignore it.