Stuck: oscillating spring question?

Question

Any hints would be good :)

A block hangs from a spring and is set into vertical oscillation with a frequency of 1.20 Hz.
Calculate the extension of the spring when the block is at the equilibrium position.

the answer is 17.2 cm, srry shoulda said that earlier

Answer 1

The equilibrium position of the spring is the position at which the net forces on the spring are zero. Normally, this is when extension is also zero. However, in the situation where a block exerts a gravitational force on the spring, the equilibrium position is instead the position where the spring force equals the force of gravity. That is, kx = mg ==> x = mg/k. You have not been given k or m, but you have been given the frequency. In an oscillating spring, f = (1/(2*pi))*sqrt(k/m). We can thus rewrite k as m*(2*pi*f)^2, and we know f = 1.2 Hz. That means k = m*(2*pi*1.2*(1/s))^2 = 56.85m*(1/s^2). Now we can say x = mg/(56.85m*(1/s^2)) = (9.8 m/s^2)/(56.85*(1/s^2)) = 0.17 m.

Answer 2

When the block is at equilibrium, the forces of the system are balanced. That is, gravitational weight of the block equals the compression force of the string.

mg = kx

g = (k/m)x

g = (w^2)x

x = g/(w^2) = g/{(2*pi*freq)^2} = 9.81 / ((1.20*2*pi)^2)

x = 0.1727m (or) x = 17.3 cm

That is the extension of the spring.