# find period from simple harmonic motion position

If an object suspended from a spring is displaced vertically from its equilibrium position by a small amount and re- leased, and if the air resistance and the mass of the spring are ignored, then the resulting oscillation of the object is called simple harmonic motion. Under appropriate condi- tions the displacement y from equilibrium in terms of time t is given by $$y=A\cos \omega t$$ where A is the initial displacement at time t = 0, and ω is a constant that depends on the mass of the object and the stiffness of the spring (see the accompanying ﬁgure). The constant |A| is called the amplitude of the motion and ω the angular frequency..... The period T is the time required to make one complete oscillation. Show that T = 2π/ω.

If I take amplitude equal to $$2\pi$$ than, I found an equation which is nearly related to period. But, I noticed there's a $$\cos$$. I don't know how to remove it. How to solve it? I would request for hint.

• How are you taking the amplitude equal to $2\pi$? Could you please elaborate about it in the question? May 29, 2021 at 16:07

Time period is defined as the time interval taken for the system to finish one complete oscillation. This means that the phase difference between the initial and final position is $$2\pi$$. This means that: $$\omega \Delta t=2\pi$$ so that $$\Delta t=\frac {2\pi}{\omega}$$ Remember that, amplitude is not a property of a system, it depends upon the external forces acting upon the system. Hence you cannot simply define it to be $$2\pi$$.
• But, what about $\cos$? How we can remove it? Where does it go?
• It doesn't "go" anywhere. Just that the phase changes by $2\pi$, since $\cos (\theta)=\cos(\theta+ 2\pi)$ May 29, 2021 at 16:27