Question

On a horizontal frictionless surface a mass M is attached to two light elastic strings both having length l and both made of the same material. The mass is displaced by a small displacement Δy such that equal tensions T exist in the two strings, as shown in the figure. The mass is released and begins to oscillate back and forth. Assume that the displacement is small enough so that the tensions do not change appreciably. (a) Show that the restoring force on the mass can be given by F = -(2T∆y)/l (for small angles) (b) Derive an expression for the frequency of oscillation.

Answer 1

ω = √(2T / (mL))

Step-by-step explanation:

(a) Draw a free body diagram of the mass. There are two tension forces, one pulling down and left, the other pulling down and right.

The x-components of the tension forces cancel each other out, so the net force is in the y direction:

∑F = -2T sin θ, where θ is the angle from the horizontal.

For small angles, sin θ ≈ tan θ.

∑F = -2T tan θ

∑F = -2T (Δy / L)

(b) For a spring, the restoring force is F = -kx, and the frequency is ω = √(k/m). (This is derived by solving a second order differential equation.)

In this case, k = 2T/L, so the frequency is:

ω = √((2T/L) / m)

ω = √(2T / (mL))