Physics Asked on November 26, 2020
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:
$$boxed{T=2pi sqrt{frac{m}{k}}}$$
So the weight of the bob $mmathbf{g}$ has no effect on it, only its actual mass $m$.
It would be different if it was a swinging pendulum.
So, consider the Newtonian equation of motion for a mass oscillating in the $y$-direction:
$$F_{net}=ma_y=mfrac{text{d}^2y}{text{d}t^2}=-ky+mg$$ $$my''(t)+ky(t)-mg=0tag{1}$$ Substitute: $$u=ky-mg$$ so: $$u'=ky'$$ and: $$u''=ky''Rightarrow y''=frac{u''}{k}$$ Insert into $(1)$: $$frac{mu''}{k}+u=0Rightarrow u''+frac{k}{m}u=0$$ 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: $$u(t)=Acos(omega t+varphi)$$ Where $A$ is the amplitude and $varphi$ the phase angle. The initial conditions are: $$y(0)=Atext{ and }y'(0)=0$$ Back-substituting we get:
$$y(t)=frac{1}{k}Big(Acos(omega t+varphi)-mgBig)$$ Using the first initial condition we can now show that: $$A=frac{1}{k}(Acosvarphi-mg)$$ $$kA=Acosvarphi-mg$$ $$(k-cosvarphi)A=-mg$$ $$A=frac{-mg}{k-cosvarphi}$$ So we can write: $$y(t)=-frac{mg}{k(k-cosvarphi)}cos(omega t+varphi)-frac{mg}{k}$$ (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: $$|y'(t)|=|frac{mgomega}{k(k-cosvarphi)}sin(omega t+varphi)|$$ This reaches a maximum for $(omega t+varphi)=nfrac{pi}{2}$ with $n=1,3,5,...$ because then the $sin$ becomes $1$.
Or: $$boxed{|y'(t)|_{max}=frac{mgomega}{k(k-cosvarphi)}}$$ So the maximum velocity does indeed depend on $g$.
In a simple sense this is quite easy to understand. For a generic oscillator: $$y(t)=Acos(omega t+varphi)$$ the maximum velocity is in fact $omega A$. It so happens that in our case $A$ is dependent on $g$.
Answered by Gert on November 26, 2020
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