A mass placed between 2 springs fixed at walls

Tuesday, 10 November 2009 15:03 Bruno Lebon

Another oscillation problem from a physics forum.

A box of mass M = 2 kg is held to two opposite walls by two springs each of spring constant k = 115 N m^-1. For simplicity, you may assume that when the box is halfway between the two walls, both springs are relaxed.

a.) With what frequency does it oscillate?

b.) Suppose now we change the problem so that the left and right springs have spring constants k₁ = 161 N m^-1 and k2 = 92 N m^-1, respectively. Now what is the frequency of oscillation?

If $x=0$ when $M$ is midway between the walls, any displacement $x$ will compress one spring by $x$ and extend the other by the same amount $x$ .

We again consider the Lagragian $L=\frac{1}{2}Mv^{2}-\frac{1}{2}k_{1}x^{2}-\frac{1}{2}k_{2}x^{2}$ .

The equation of motion becomes $Ma+\left(k_{1}+k_{2}\right)x=0$ . Again, the solution to this equation is of the form $x(t)=x_{0}\cos\left(\omega t+\phi\right)$ where $x_{0}$ is the amplitude of oscillation, $\omega$ the angular frequency and $\phi$ the initial phase.

Obviously, when solving the equation of motion, we obtain $\omega=\sqrt{\left(k_{1}+k_{2}\right)M}$ and we can easily compute the value of the frequency using $\omega=2\pi f$ .