# Does maximum velocity change when vertical mass-spring system is used in different location on Earth in SHM?

Let me elaborate for you my concerning

I am thinking of a example of a vertical mass spring system. Suppose i place my system at equator, let suppose a wall clock which uses a vertical spring mass system to measure time. Each time the mass reaches an extreme position, the clock advanced to by a second, so 1 sec is our time period. Because of having some weight (mg'), our equilibrium position shift a little bit (say X1)

Now i place my system at poles as we knows that due to earth rotation gravity increases at poles so, now because of having some weight (mg), our equilibrium position shifts a little bit (say X2)

Where g > g' and X2 > X1

Coming to the question

what is the maximum velocity in both case? Is both are equal? Or different?

The period of oscillation of a mass-spring system is given by:

$$\boxed{T=2\pi \sqrt{\frac{m}{k}}}$$

So the weight of the bob $$m\mathbf{g}$$ has no effect on it, only its actual mass $$m$$.

It would be different if it was a swinging pendulum.

To answer the central question the OP asked, we need to solve the Newtonian equation of motion. Armed with that knowledge we'll be able to calculate the maximum velocity.

So, consider the Newtonian equation of motion for a mass oscillating in the $$y$$-direction:

$$F_{net}=ma_y=m\frac{\text{d}^2y}{\text{d}t^2}=-ky+mg$$ $$my''(t)+ky(t)-mg=0\tag{1}$$ Substitute: $$u=ky-mg$$ so: $$u'=ky'$$ and: $$u''=ky''\Rightarrow y''=\frac{u''}{k}$$ Insert into $$(1)$$: $$\frac{mu''}{k}+u=0\Rightarrow u''+\frac{k}{m}u=0$$ Now substitute: $$\omega^2=\frac{k}{m}$$ So that: $$u''(t)+\omega^2u(t)=0$$ Which is the classical diff. equation of a harmonic oscillator and has the solution: $$u(t)=A\cos(\omega t+\varphi)$$ Where $$A$$ is the amplitude and $$\varphi$$ the phase angle. The initial conditions are: $$y(0)=A\text{ and }y'(0)=0$$ Back-substituting we get:

$$y(t)=\frac{1}{k}\Big(A\cos(\omega t+\varphi)-mg\Big)$$ Using the first initial condition we can now show that: $$A=\frac{1}{k}(A\cos\varphi-mg)$$ $$kA=A\cos\varphi-mg$$ $$(k-\cos\varphi)A=-mg$$ $$A=\frac{-mg}{k-\cos\varphi}$$ So we can write: $$y(t)=-\frac{mg}{k(k-\cos\varphi)}\cos(\omega t+\varphi)-\frac{mg}{k}$$ (We don't need to determine the value of $$\varphi$$ here)

The velocity (here as absolute value) is given by:

$$|y'(t)|=|\frac{\text{d}y(t)}{\text{d}t}|$$ Which gives: $$|y'(t)|=|\frac{mg\omega}{k(k-\cos\varphi)}\sin(\omega t+\varphi)|$$ This reaches a maximum for $$(\omega t+\varphi)=n\frac{\pi}{2}$$ with $$n=1,3,5,...$$ because then the $$\sin$$ becomes $$1$$.

Or: $$\boxed{|y'(t)|_{max}=\frac{mg\omega}{k(k-\cos\varphi)}}$$ So the maximum velocity does indeed depend on $$g$$.

In a simple sense this is quite easy to understand. For a generic oscillator: $$y(t)=A\cos(\omega t+\varphi)$$ the maximum velocity is in fact $$\omega A$$. It so happens that in our case $$A$$ is dependent on $$g$$.

• But What about its Max Velocity in DiFferent cases – 5 Dots Sep 19 at 18:17
• You Should edit your answer as i already aware of Time Period is Same in both Cases,I want to Know about its Max Velocity at equilibrum position in different cases – 5 Dots Sep 19 at 18:23
• Sorry for first misreading your question. The answer is now correct and complete. – Gert Sep 19 at 20:44
• Please edit your answer again as in that line Insert into (1)(1): mu′′k+u=0⇒u′′=kmu=0. The correction is. u"= - kmu and u" +kmu = 0 and in that line y(t)=k(Acos(ωt+φ)−mg)/k the correction is y(t)=(Acos(ωt+φ)+mg)/k and finally how you got A (amplitude) = mg/(k-cosφ) – 5 Dots Sep 20 at 2:32
• Thanks for proof readingt Edits made. – Gert Sep 20 at 14:59