# 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

## 1 Answer

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?
– user876873
May 29, 2021 at 16:24
• 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