A spring of length $l$ and spring constant $k$ is suspended vertically with an object with mass $m$ attached at the bottom. If you take the top of the spring and oscillate it such that its displacement is $a\sin\omega t$ what is the object's equation of motion? Note, there is no damping and gravitational acceleration is $g$.

Clearly the EOM is meant to show a driven oscillation. I understand that the solution can be obtained using the Lagrangian, but since we haven't covered that in class, I cannot use it.

This was my attempt:

Let $X$ be the displacement of the object from the equilibrium point of the spring. Then, if $\omega_0=\sqrt{\frac{k}{m}}$, \begin{equation} \ddot{X}+\omega_0^2X=g \end{equation} Then, let $x$ represent the displacement of the object with the origin at the top end of the spring when $t=0$. Therefore, $x=a\sin\omega t+l+X$. Substituting this into the previous equation, \begin{equation} \ddot{x}+\omega_0x=g-\omega_0^2l+(\omega_0^2-\omega^2)a\sin\omega t \end{equation}

I am not confident about this solution since when $\omega=\omega_0$, the driven oscillation in the equation disappears. What did I do incorrectly?

  • $\begingroup$ Actually x goes to infinity as $\omega$ goes to $\omega_o$ $\endgroup$
    – docscience
    Jan 21, 2015 at 3:59

2 Answers 2


You take the equations of motion for the mass and apply a generic harmonic function. Solve for the coefficients in order to satisfy the EOM for all times

Spring force $F = -k (-\ell+x-a \sin( \omega t))$

Equations of motion $m \ddot{x} = F - m g$

Fit a general displacement function $x =C_0 + C_1 \sin(\omega t)$

Use them together to get

$$ \sin(\omega t) \left( a k - C_1 (k- m \omega^2) \right) + k (\ell-C_0)-m g =0 $$

This has solution $C_0 = \ell - \frac{m g}{k}$ and $C_1 = \frac{a k}{k-m \omega^2}$

$$\boxed{x(t) = \ell - \frac{m g}{k} + \frac{a k}{k-m \omega^2} \sin(\omega t) }$$

  • $\begingroup$ This looks correct to me, although you're using a different definition of $x$ from the one the OP defines. $\endgroup$
    – Brionius
    Jan 21, 2015 at 4:17

Let's check your answer starting back at Newton's 2nd Law:

$$F_{net} = m \ddot{x}$$

$$F_{spring} + F_{gravity} = m \ddot{x}$$ $$-k (x - x_{spring} - l) + m g = m \ddot{x}$$ $$-k (x - a sin(\omega t) - l) + m g = m \ddot{x}$$ $$-k x + a k sin(\omega t)) + k l + m g = m \ddot{x}$$ $$\ddot{x} + \frac{k}{m} x = g + \frac{k l}{m} + \frac{a k}{m} sin(\omega t))$$

Substituting $\omega_0 = \sqrt{\frac{k}{m}}$, we get our equation of motion:

$$\ddot{x} + \omega^2_0 x = g + \omega^2_0 l + a \omega^2_0 sin(\omega t))$$

To check that this is plausible, we can find solutions of the form

$$x = \alpha sin(\omega t) + \beta$$

Substituting and finding $\alpha$ and $\beta$,

$$x = \frac{a}{\frac{\omega}{\omega_0}^2 - 1} sin(\omega t) + \frac{g}{\omega^2_0} + l$$

This shows the expected behavior with respect to gravity, the length, and the resonance as $\omega$ goes to $\omega_0$.


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