Question

A 50-cm-long spring is suspended from the ceiling. A 250 g mass is connected to the end and held at rest with the spring unstretched. The mass is released and falls, stretching the spring by 20 cm before coming to rest at its lowest point. It then continues to oscillate vertically. a. What is the spring constant? b. What is the amplitude of the oscillation? c. What is the frequency of the oscillation?

Answer 1

(a) 24.5N/m

(b) 10cm (0.1m)

(c) 1.58osc/s

Step-by-step explanation:

We can obtain the spring constant using the conservation of mechanical energy; before the mass is released, it only has gravitational potential energy, and in its lowest point has only elastic potential energy. Then, we get:

`U_g_0=U_e_f\\\\mgh=(1)/(2)kh^2\\\\k=(2mg)/(h)\\\\k=(2(0.25kg)(9.8m/s^2))/(0.2m)\\\\k=24.5N/m`

Then, the spring constant is 24.5N/m (a).

Since the mass falls 20cm and conserves its energy, when it rises again it will go 20cm upwards. So, the amplitude of the oscillation is:

`A=(h)/(2) \\\\A=(20cm)/(2)\\ \\A=10cm`

The amplitude is 10 cm (or 0.1m) (b).

Now, the frequency of a mass-spring system is given by:

`f=(1)/(2\pi)\sqrt{(k)/(m)}`

And, plugging in the given values, we get:

`f=(1)/(2\pi)\sqrt{(24.5N/m)/(0.25kg)}\\\\f=1.58osc/s`

It means that the frequency of the oscillation is 1.58osc/s (c).