# Find the frequency of oscillation of two masses connected by a spring [duplicate]

Question:

Two masses $$m_1$$ and $$m_2$$, constrained to the $$x$$-axis, are coupled by a light spring of stiffness $$s$$ and natural length $$l$$. If $$x$$ is the extension of the spring and $$x_1$$ and $$x_2$$ indicate the positions of the two masses ($$x_2 > x_1$$), show that the equations of motion of the masses along the $$x$$-axis are:

$$m_1\ddot{x}_1 = sx$$ and $$m_2\ddot{x}_2 = −sx$$

and combine these to show that the system oscillates with a frequency:

$$ω = \sqrt{\frac{s}{\mu}}$$

where $$\mu = \frac{m_1m_2}{m_1 + m_2}$$ is the reduced mass.

Attempt: So I know $$F = ma = -kx$$ and acceleration is the second derivative of $$x$$ so $$a_1 = \ddot{x}_1$$ and $$a_2 = \ddot{x}_2$$. So that gives the first two equations: $$m_1\ddot{x}_1 = sx$$ and $$m_2\ddot{x}_2 = −sx$$ and they have different signs because they the force acts in opposite directions.

Then I know the extension of the spring $$x = x_2 - x_1 - l$$, so $$m_1\ddot{x}_1 = s(x_2 - x_1 - l)$$ and $$m_2\ddot{x}_2 = -s(x_2 - x_1 - l)$$, but then combining these all I seem to get is $$m_1\ddot{x}_1 + m_2\ddot{x}_2 = 0$$, so I'm not sure where to go from there?

So any help would be appreciated.

• Oct 9 at 21:47

## 1 Answer

You have found the two differential equations \begin{align} m_1\ddot{x}_1&=s(x_2-x_1-l) \\ m_2\ddot{x}_2&=-s(x_2-x_1-l) \end{align} \tag{1} for the unknown functions $$x_1(t)$$ and $$x_2(t)$$.

Because you already guess the solution will be an oscillatory motion, you can make the approach \begin{align} x_1(t)&=A_1\sin(\omega t) \\ x_2(t)&=l+A_2\sin(\omega t) \end{align} \tag{2} where $$A_1$$, $$A_2$$ and $$\omega$$ are some still unknown constants.

Inserting the approach (2) into the differential equations (1) you get \begin{align} -m_1A_1\omega^2&=s(A_2-A_1) \\ -m_2A_2\omega^2&=-s(A_2-A_1) \end{align} \tag{3}

There are many different ways (basic and sophisticated) to solve this. A basic way is:

• resolve one equation of (3) to $$A_2$$,
• insert this $$A_2$$ into the other equation of (3),
• resolve the resulting equation for $$\omega^2$$.

You will find the result to be $$\omega^2=s\left(\frac{1}{m_1}+\frac{1}{m_2}\right)$$ and the constant $$A_1$$ will still be arbitrary.