📂Classical Mechanics

Multiple Spring Oscillation

When springs are connected on both sides of an object

Let $x$ be the distance the object has moved. Since the restoring force of the spring is $-kx$, the object receives a force of $-k_{1}x$ from the left spring and $-k_{2}x$ from the right spring. Therefore, the equation of motion is as follows.

$$\begin{align*} && m\ddot{x}&=-k_{1}x-k_{2}x \\ \implies &&m\ddot{x}+(k_{1}+k_{2})x&=0 \\ \implies && \ddot{x}+\frac{k_{1}+k_{2}}{m}x &=0 \end{align*}$$

This is the same as the equation for simple harmonic motion, so the solution is as follows.

$$\begin{align*} x(t) &= A\cos(\omega_{p} t + \phi) \end{align*}$$

Here, $A$ is the amplitude and $\omega_{p} =\sqrt{\frac{k_{1}+k_{2}}{m}}$ is the frequency. In other words, it’s like adding the two spring constants in the solution for simple harmonic motion.

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When different springs are connected together

Let $x_{1}$ be the stretched length of spring 1, and $x_{2}$ be the stretched length of spring 2. According to the law of action-reaction, the force exerted by spring 1 on spring 2 is equal in magnitude to the force exerted by spring 2 on spring 1. Therefore, the following equation is obtained.

$$\begin{equation} \left| F_{12} \right| =\left| F_{21} \right| \quad \implies \quad k_{1}x_{1}=k_{2}x_{2} \label{eq1} \end{equation}$$

At this point, the distance the object has moved is $x=x_{1}+x_{2}$. If we consider the combined springs as one spring with a spring constant of $k$, then the equation of motion is as follows.

$$\begin{equation} F = -kx \label{eq2} \end{equation}$$

In this case, the force exerted by spring 1 is canceled out by the force exerted by spring 2 on spring 1, so the net force is $F=-k_{2}x_{2}$. Then, by $\eqref{eq1}$ and $\eqref{eq2}$, it is as follows.

$$\left| F \right|=kx=k_{1}x_{1}=k_{2}x_{2}$$

Therefore, the following equation holds.

$$\begin{align*} && x&=x_{1}+x_{2} \\ \implies && \frac{\left| F \right| }{k} &=\frac{ \left| F \right| }{k_{1}}+\frac{\left| F \right| }{k_{2}} \\ \implies && \frac{1}{k} &=\frac{1}{k_{1}}+\frac{1 }{k_{2}} \\ \implies&& k&=\frac{k_{1}k_{2}}{k_{1}+k_{2}} \end{align*}$$

Thus, if we rewrite $\eqref{eq2}$, it becomes as follows. $$F=-kx=-\frac{k_{1}k_{2}}{k_{1}+k_{2}}x$$

If we set $\omega_{s}^{2}=\frac{k_{1}k_{2}}{k_{1}+k_{2}}$, the solution of the equation of motion is given as follows.

$$x(t) = A \cos (\omega_{s} t +\phi)$$

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See also

• Simple Harmonic Motion
• Damped Oscillation
• Forced Oscillation
• Coupled Oscillation