- #1
Je m'appelle
- 117
- 0
I need someone to please verify my work.
A particle of mass m is suspended from the ceiling by a spring of constant k and initially relaxed length l_0. The particle is then let go from rest with the spring initially relaxed. Taking the z-axis as vertically oriented downwards with it's origin in the ceiling, calculate the position 'z' of the particle as a function of time 't'.
SHO equation:
m\ddot{x}(t) = -kx(t)
In this case, the acting forces on the mass are the weight directed downwards and the elastic force of the spring directed upwards, so
m\ddot{x}(t) = mg - kx(t)
\ddot{x}(t) = g - \omega^2 x(t)
\ddot{x}(t) + \omega^2 x(t) = g
It's important to note that I'm assuming 'x' as the displacement of the spring and NOT the displacement of the mass.
\ddot{x}(t) + \omega^2 x(t) = g
x(t) = x_p(t) + x_h(t)
The homogeneous solution will be
x_h(t) = acos(\omega t) + bsin(\omega t)
And the particular solution will be
x_p(t) = \frac{g}{\omega^2}
By using the initial conditions x(0) = 0, \dot{x}(0) = 0 we get
x(0) = 0 = a + \frac{g}{\omega^2}
a = - \frac{g}{\omega^2}
\dot{x}(0) = 0 = b\omega,\ b = 0
So by rearranging we get to
x(t) = -\frac{g}{\omega^2}cos(\omega t) + \frac{g}{\omega^2}
But notice that this is the displacement of the spring and not of the mass, in order to find the displacement of the mass we would have to add the initial length of the relaxed spring, that is, l_0.
So the final answer for the displacement of the mass in the z-axis will be
z(t) = l_0 - \frac{mg}{k} (cos(\sqrt{\frac{k}{m}} t) - 1)
Could someone please verify this, as there is a different answer in a solution manual I found, and I'm not so sure which one is correct.
Homework Statement
A particle of mass m is suspended from the ceiling by a spring of constant k and initially relaxed length l_0. The particle is then let go from rest with the spring initially relaxed. Taking the z-axis as vertically oriented downwards with it's origin in the ceiling, calculate the position 'z' of the particle as a function of time 't'.
Homework Equations
SHO equation:
m\ddot{x}(t) = -kx(t)
The Attempt at a Solution
In this case, the acting forces on the mass are the weight directed downwards and the elastic force of the spring directed upwards, so
m\ddot{x}(t) = mg - kx(t)
\ddot{x}(t) = g - \omega^2 x(t)
\ddot{x}(t) + \omega^2 x(t) = g
It's important to note that I'm assuming 'x' as the displacement of the spring and NOT the displacement of the mass.
\ddot{x}(t) + \omega^2 x(t) = g
x(t) = x_p(t) + x_h(t)
The homogeneous solution will be
x_h(t) = acos(\omega t) + bsin(\omega t)
And the particular solution will be
x_p(t) = \frac{g}{\omega^2}
By using the initial conditions x(0) = 0, \dot{x}(0) = 0 we get
x(0) = 0 = a + \frac{g}{\omega^2}
a = - \frac{g}{\omega^2}
\dot{x}(0) = 0 = b\omega,\ b = 0
So by rearranging we get to
x(t) = -\frac{g}{\omega^2}cos(\omega t) + \frac{g}{\omega^2}
But notice that this is the displacement of the spring and not of the mass, in order to find the displacement of the mass we would have to add the initial length of the relaxed spring, that is, l_0.
So the final answer for the displacement of the mass in the z-axis will be
z(t) = l_0 - \frac{mg}{k} (cos(\sqrt{\frac{k}{m}} t) - 1)
Could someone please verify this, as there is a different answer in a solution manual I found, and I'm not so sure which one is correct.
(just kidding of course)
