# Oxford PAT 2014, Question 14

## 9 thoughts on “Oxford PAT 2014, Question 14”

1. Hey there.

I am not quite sure about (ma=-kx). For vertical spring oscillation, isn’t ma=-(kx-mg) ? since at any given position during oscillation the mg force is experienced? Thus, this is not an SHM, is it? Also, if period is to be found, doesn’t g now influence the period experienced, since according to ma=-(kx-mg) the larger g implies smaller net acceleration and thus larger period?

1. You’re right that gravity acts on the mass and tends to pull it down. But we can make our job simpler by putting the point x = 0 at the equilibrium position, where the spring is extended such that the force mg downwards is balanced by an upwards force caused by the extension of the spring. So, in the equation I have written, x = 0 is the position that the mass would have when not moving or accelerating, because the extension of the spring is pulling the mass up just as strongly as gravity is pulling the mass down.

2. Clary says:

Can you explain on how the value of g will affect the position of equilibrium point?

1. If the spring constant is k, and the mass is m, then the equilibrium point is distance x below the bottom of the un-extended spring, where kx = mg (i.e. the upwards force exerted by the spring cancels out the downwards force of gravity acting on the mass. So as g increase the equilibrium point moves lower.

1. Clary says:

There’s another formula where T=2pi (square root) (l/g). Is that the formula for period of the pendulum? So in other word, the g will not affect the period of oscillation on the spring but there’s an effect on the period of pendulum?

1. That’s right. You should be able to see why this is if you draw diagrams for the spring case and the pendulum case, and consider the respective forces.

Simple harmonic motion is what you get when you have some restorative force that is proportional to the quantity that is varying. In the case of the mass/spring, the varying quantity is the distance x of the mass below the equilibrium point, and the restorative force is kx. In the case of the pendulum, the varying quantity is the (small) angle of deflection theta and the restorative force is mg sin(theta) which is mg theta when theta is small. For the mass/spring, the restorative force does not depend on g, and that’s why the period is independent of g. Incidentally, in a similar vein it’s also worth noting that the acceleration of the pendulum back towards the equilibrium point is mg theta / m, which equals g theta with the m cancelling out, so the period of a pendulum is independent of m.

3. Harry says:

could you please explain the first bit- why does ma=-kx especially the minus bit and where does sine come into this and how does this determine the period? thanks

1. Hi Harry. This is an example of simple harmonic motion, which I don’t think is covered at AS level, so you probably won’t have heard of it yet. There is a link here, but I am sure there must be good tutorial links elsewhere.

Here is an explanation of the ma = -kx bit. The value k is the spring constant, and x is the distance of the mass below the equilibrium position. When the mass is in the equilibrium position, x is zero, the spring is extended but only enough to balance out the (constant) gravitational force on the mass. As the mass moves down, x increases and so there is an additional upwards force of kx; because there is an upwards force of kx there is an upwards acceleration, which is of course the same thing as a negative downwards acceleration. That is where you get the minus sign.

Where does sine come in? The formula ma = -kx is a well-known differential equation, and x = sin(t sqrt(k/m)) is a solution to this equation a = -(k/m)x. If you differentiate the sin(t sqrt(k/m)) with respect to t: you get (sqrt(k/m))cos(t sqrt(k/m)), and if you differentiate it again you get (sqrt(k/m))(sqrt(k/m))-sin(t sqrt(k/m)), which is -(k/m)x.

1. Harry says:

thanks