Simple Harmonic Motion Test Question

Jackel

I've been set this question by my tutor and I'm having difficulty doing it.

A mass of 2 kg is hung from the lower end of a vertical spring and extends it by 40 cm. The mass is now pulled down a further 20 cm and is then released from rest so that it oscillates about the equilibrium position. Determine :

a) the spring stiffness constant k for the spring.
b) the time period of the oscillations.
c) the speed and acceleration of the mass when it is 15 cm from the equilibrium position.
d ) the maximum speed and maximum kinetic energy of the mass.
e) the maximum accelerating force on the mass.

I'm struggling with it as I don't really understand how to find the K. Is the displacement 20? Or would it be 60? I think K is, F/x, x being the displacement?

If someone could please explain it to me I would be very grateful. I've been out of education for some time so I'm finding some things hard to get my head around.

ozone

The distance that you would use in your k calculation is the distance which the spring has moved from equilibrium. Dont forget to multiply the weight by 9.8 to find the force in newtons

Jackel

Right so for K I have, 981.

F = M.A, 2kg x 9.81 = 19620 N

K = F/x, 19620/20 = 981.

Is that right?

ozone

Well let me put it this way. If the spring is hanging and has no weight on it, then wouldn't it be in equilibrium?

Jackel

Oh, so equilibrium would be 0cm? As the weight pulls it out of equilibrium, so then the displacement would be 60cm?

ozone

Close, but I think that you are to assume that work from outside the mass/spring system is pulling the weight down that last 20 cm... This displacement will most likely set the spring into an oscillatory motion.

Jackel

Right, so the equilibrium position is 0 and the displacement is 40? So the extra 20cm is used to solve the other questions? It's not used in the string stiffness question?

I think I'm starting to confuse myself now, lol.

PeterO

I beg to differ with your implications.

The equilibrium position is clearly where the mass would be if not oscillating - ie with the spring extended 40cm.
Why didn't you take the equilibrium position to be when the spring was in the drawer, in the store-room, with all the other springs?

The weight of the mass extends the spring by 40 cm to create that equilibrium position.

Some one/thing THEN extends the spring a further 20cm, from where the mass oscillates.

While oscillating, it will move between extension 20cm and extension 60cm [ie either side of the equilbrium position 40cm-extension]

If you take the equilibrium position as extension zero, it it is very difficult to find "the speed and acceleration of the mass when it is 15 cm from the equilibrium position." as it would never gets to a position 15cm from zero.
It does of course pass through a couple of points 15cm from the 40cm equilibrium position.

PeterO

I am not surprised - read my previous post.

sirisha kotik

1) kx=mg where x=40 cm.
2) time period wud be 2pi(m/k)^1/2
3) equilibrium position wud be wher kx=mg. i.e. 40 cm. n all the other parts cn be solved by using energy conservation.

Jackel

Yeah, that makes more sense. Thanks!