How do you prove that maximum speed occurs at the center of oscillation?
In VCAA 2014 Q2(d), they've stated that maximum speed occurs at the centre of oscillation in the examiners report, did we have to know this as a part of our 'prerequisite' knowledge or was there a way to work this out?
Hmmm... on the exam I got stuck on that q because I didn't want to assert that KE(max) was at the centre of oscillation - I know it's "intuitive" but I wasn't sure I wanted to risk marks and make that assumption.
So I turned it into a methods question, found a quadratic for KE in terms of x and got the maximum (tick)
But that was a crazy method of solving the question ... did anyone feel like the question was a bit fuzzy? After looking at the graphs and thinking about it over and over, I felt that it was possible that the maximum could occur NOT at the centre of oscillation, because SPE was quadratic and GPE was linear, thus SPE+ GPE could minimise at some weird place not at the centre ...
Hopefully I'll get better at this.
Does anyone know if you'd be penalised for using this sort of knowledge to solve the problem? Totally valid but using methods techniques.
Let me mathematically convince you that the centre of oscillation necessarily is the point of maximum KE.
I'll do this two ways.
First way: using forces.
The centre of oscillation is the point of zero net force, so you have F = ma = 0. This means that you have zero acceleration here, so the velocity must be stationary. Now, F = -kx - mg. As F is linear in x, it changes sign at the stationary point. That means on one side of the centre of oscillation the velocity is increasing and on the other side it's decreasing.
If you increase x, F decreases so when x gets higher, the force is negative. If you decrease x, F increases so when x gets lower, the force is positive. The net effect is that as the force opposes any displacement from equilibrium, your speed will decrease either side of the centre of oscillation. So the centre of oscillation gives the maximum KE.
Second way: using energy
You have 1/2 mv^2 + 1/2 kx^2 + mgx = constant
Let's find the centre of oscillation explicitly. From above. 0 = mg + kx -> x = -mg/k. Remember this result.
Now, let's complete the square on the above.
1/2 mv^2 + 1/2 k(x^2 + 2mg/k x) = 1/2 mv^2 + k/2 ((x+mg/k)^2 - some constant) = constant
I haven't written out the constant because the exact values of these constants aren't important.
1/2 mv^2 + 1/2 k(x+mg/k)^2 = constant
So this is an ellipse. Can you see that the speed is greatest when x = -mg/k, aka the centre of oscillation?