- #1
frozen7
- 163
- 0
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.
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?
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?