Question

a 400 gram mass is attached to a spring with a 300 n/m spring constant. what is the angular frequency of this system?

Answer 1

Let the periodic motion of the mass attached to the spring be represented by x = Ａsin(ωt), where x is the position of the mass, A is the amplitude of the motion and ω the angular frequency.

In order to obtain the velocity, we differentiate x.

v = dx/dt = d[Ａsin(ωt)] / dt = Ａd[sin(ωt)] / dt = Ａω cos(ωt)

In order to obtain the accelaration, we differentiate v.

a = dv/dt = d[Ａωcos(ωt)] / dt = Ａωd[cos(ωt)] / dt = -Ａω² sin(ωt)

Force experienced by the mass = ma = -mＡω² sin(ωt) = -mω²[Ａsin(ωt)] = -mω²x ---------------------- (1)

Now, the force experienced by a mass attached to a spring when it is displaced by a distance x is F = -kx, where k is the spring constant. Equating this with (1):

-mω²x = -kx ⇒ mω² = k ⇒ ω = (k/m)^1/2

k = 300 N/m; m = 0.4Kg

Thus, ω = (300/0.4)^1/2 = (750)^1/2 = 27.38 s^-1

The answer is 27.38 s^-1