In a vertical mass spring system, spring with spring constant 250 N/m vibrates with an amplitude of 12cm when 0.38kg hangs from it.(adsbygoogle = window.adsbygoogle || []).push({});

What is the equation describing this motion as a function of time? ( assume the mass passes through the equilibrium point, towards the positive x(upward), at t = 0.110s)

I do it in this way:

Hooke`s Law: F = kx

(0.38)(9.8)= 250 N/m (x)

x = 0.015(m)

[tex]\omega[/tex] = [tex]\sqrt{k/m}[/tex]

[tex]\omega[/tex] = 25.65

x = Acos([tex]\omega[/tex]t + [tex]\phi[/tex])

0.015 = 0.12cos(25.65t + [tex]\phi[/tex])

[tex]\phi[/tex] = -1.38

So, the equation : x = 0.12cos(25.65t -1.38)

Is it correct?

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# Homework Help: Mass Spring System

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