Assume you are looking at a top view of an object of mass m connected between 2 stretched rubber bands of length L. The objects rests on a frictionless surface. At equilibrium, the tension in each rubber band is T. Find an expression for the frequency of oscillations perpendicular to the rubber bands. Assume the amplitude is sufficiently small that the magnitude of the tension in the rubber bands is essentially unchanged as the mass oscillaties.
The answer is: f = (1/2pi)*Sqrt(2T/mL)
It helps to draw a picture of the object at some position during the oscillation. If you do this then don't draw the case of a very small amplitude, because then you don't see the important things in your drawing! So, just draw a picture representing the object at some large deviation.
If you do that, then what you see is that the two rubber bands are no longer parallel. The two rubber bands and the object form two sides of a triangle. Now, let's draw a line from the object at ...
A detailed solution is given.