A spring with spring constant k is attached to mass M. The spring is horizontal

Question

A spring with spring constant k is attached to mass M. The spring is horizontal and mass M sits on a horizontal frictionless table. On the side of mass M opposite to the spring a cord is attached that runs over a pulley (which you may take to be ideal: zero mass, zero moment of inertia, and frictionless). The cord is attached to a hanging mass m.

1) Find the equilibrium position of the masses. (That means: How far does the spring stretch from its un-stretched length?)
2) Now consider small oscillations about this equilibrium position. What is their angular frequency (!)?

Thank you :)

Explanation / Answer

1. The spring stretches by mg/k (small m, mass of hanging object). Just draw the free-body diagram, and let the tension = T in the cord. T - kx=0 for mass M and T-mg =0 for mass m in equilibrium. Eliminate T to get the result.
2. The accelerations of the two masses are the same at any instant, because the cord between the two masses cannot be stretched or compressed. This is a trivial example of what's called a kinematic constraint. Write down Newton's law for both masses:

mg - T = ma = mx'' where '' denotes second derivative.

T - kx= Ma = Mx''. Eliminate T to get:

(m+M)x'' + kx - mg = 0. This is the equation for SHM with an "angular" frequency k/(m+M).

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