# Simple pendulum in elevator

1. Jul 20, 2014

### Tanya Sharma

1. The problem statement, all variables and given/known data

A simple pendulum of length l is tied to the ceiling of an elevator which is at rest. The pendulum is oscillating with a time period T, and it has an angular amplitude β. Now at some time when the bob of the pendulum is at the mean position, the elevator suddenly starts moving down with acceleration a. What is the new angular amplitude of the pendulum?

2. Relevant equations

3. The attempt at a solution

I tried a few things but haven’t been able to come with something useful.

When the elevator is at rest the governing equation is $\frac{d^2θ}{dt^2} = -\frac{g}{L}θ$ .

When the elevator accelerates down the governing equation is $\frac{d^2θ}{dt^2} = -\frac{(g-a)}{L}θ$.

This doesn't help much .

When thinking about energy conservation , I think energy is not conserved between the times when the elevator is at rest and when it accelerates down.

I would be grateful if somebody could help me with the problem .

2. Jul 20, 2014

### Simon Bridge

Do you know the equation for the period of a simple pendulum?
Have you tried solving the governing equation for an accelerating elevator?

Have you recently had some lessons about doing physics in a non-inertial frame?

3. Jul 20, 2014

### BruceW

well said. So what do we know? You know everything about the pendulum, just before the acceleration happens. So maybe you can use this, to work out everything about the pendulum after the acceleration.

4. Jul 20, 2014

### ehild

You are in an accelerating lift. Do you feel heavier or lighter? The pendulum "feels" the same - a different gravitational constant as before.

The pendulum was at the equilibrium position, moving with maximum speed. Is that speed different at the instant when the lift starts to accelerate?

How the maximum speed at the lowest position is related to the angular amplitude?

By the way, the governing equation for the pendulum is $\frac{d^2θ}{dt^2} = -\frac{g}{L}\sin(θ)$.

ehild

5. Jul 20, 2014

### Tanya Sharma

Hello ehild

Thanks for the response .

lighter

No . Same speed .

$v_{max}=ωβL$

I had used small angle approximation .

Is the answer $$β' = β\sqrt{\frac{g}{g-a}}$$ ?

6. Jul 20, 2014

### ehild

It is correct if the angular amplitude is very small. But the problem text does not state so.

ehild

7. Jul 20, 2014

### Tanya Sharma

OK .

1) What is your opinion regarding whether energy is conserved or not between the two stages ?

2) How would we approach the problem if the elevator had started accelerating when the bob was at some position other than mean position for example at angle $0<\theta<\beta$?

Last edited: Jul 20, 2014
8. Jul 20, 2014

### Staff: Mentor

There is no need to consider non-inertial frames, changing the effective gravitational acceleration is easier.

Did you calculate the energy before and after the elevator started?
Be careful: this depends on your choice where "zero height" is.

In the same way: both position and velocity do not change, but the motion afterwards will differ.

9. Jul 20, 2014

### Tanya Sharma

If I consider the bottommost position of the bob to be h=0 .

When the elevator is stationary $E_1 = mgL(1-cosβ)$

When the elevator starts accelerating downwards $E_2 = m(g-a)L(1-cosβ')$ , where $β' = β\sqrt{\frac{g}{g-a}}$

$E1 ≠ E2$

Does that mean energy is not conserved between the two stages ?

Last edited: Jul 20, 2014
10. Jul 20, 2014

### Staff: Mentor

Your formula for β' assumes small angles, the other formula does not. You cannot mix them like that.

It is easier to compare the current energy as sum of potential and kinetic energy, without using the maximal amplitude.

11. Jul 20, 2014

### Tanya Sharma

You are right .

The kinetic energy at the bottommost point is equal in magnitude to the total energy . Since the KE is same at the bottommost point , then the total energy is same between the two stages .

So the energy is conserved between the two stages .

Is that correct ?

12. Jul 20, 2014

### BruceW

I'd agree there isn't a discontinuous change in energy between the two stages. But this doesn't mean that energy is conserved once you're in the second stage. However, you do have the equation
$$\frac{d^2θ}{dt^2} = -\frac{(g-a)}{L} \sin(θ)$$
So you can create some kind of conserved quantity 'Q' from this equation, even though Q is not the same as the energy of the falling pendulum. (and hopefully Q will let you say something about the max angle of the pendulum).

13. Jul 20, 2014

### ehild

The inertial force F=ma has potential, so you can speak about conservation of energy: KE+PE(gravitational) + PE(inertial) = const.
We can also say that g'=g-a is the gravitational acceleration in the lift and then the change of potential energy is ΔPE = mg'Δh=m(g-a)Δh.

Remember how the Work-Energy Theorem was derived from Newton's second equation: It was a simple transformation of variable x:

$m\ddot x =F$ is the same as $m\dot v= F$ We choose x as independent variable:
$dv/dt=(dv/dx) (dx/dt) = (dv/dx) v = 1/2 d(v^2)/dx$. So we have the equation $1/2 md(v^2)/dx=F$. Integrating with respect to x it is the Work-Energy Theorem: $1/2 mv^2-1/2 mv_0^2=\int {Fdx}= W$ and in case of conservative forces, W =-PE.
You can do the same here. dθ/dt=ω and d2θ/dt2= ω(dω/dθ) .

ehild

14. Jul 21, 2014

### Tanya Sharma

ehild... - Energy changes between the two stages if elevator accelerates when the bob is at mean position and energy remains same between the two stages if bob is at some position other than mean position .

Do you agree with this ?

Last edited: Jul 21, 2014
15. Jul 21, 2014

### ehild

I agree with mfb. Both the velocity and the height of the bob is the same just before switching to acceleration and just after it.
The lift in rest and the accelerating lift are two different systems, with two different g-s. The potential energy of the bob in the lift in rest is different from that in the accelerating lift if you set the zero of the PE at the lowest position of the bob and define PE as mgh with the appropriate 'g' in both systems.
The potential energy is determined by an additive constant, so you can choose them equal in both systems initially.
There is no sense to speak about change of potential energy between the two stages which means the difference of PE in two different systems. But the energy does not change if seeing from the ground. Imagine you are a butterfly, sitting on a flower on the ground, and see a galloping horse. You know the mass and velocity and say that the horse has got 1/2 Mv2 energy.
Then you fly onto the back of the horse, and you observe that it has zero velocity (with respect to you). So its kinetic energy became zero.
Has it sense saying that the energy of the horse changed between the two stages? :tongue:

The energy of the bob is conserved in the accelerating frame of reference. We only calculate the PE with different g.

ehild

Last edited: Jul 21, 2014
16. Jul 21, 2014

### Tanya Sharma

OK.

I see your point .Talking about energy changes between different reference frames is not a good idea .

I have been thinking about this because if we consider same total energy between the two stages and assume small angle approximation , we would end up with same result as I had obtained in post#5 using maximum speed of the bob.

So this made me think what if the question had instead stated that elevator accelerates down when the bob is not at the mean position .

1) Without using small angle approximation , how would we solve the OP ?

2) How would we approach the problem if the question had instead stated that elevator accelerates down when the bob is not at the mean position i.e $0<\theta<\beta$

Last edited: Jul 21, 2014
17. Jul 21, 2014

### BruceW

yeah, good. But as ehild says, since you have switched your frame of reference, it doesn't really make sense to discuss whether energy is conserved between the two different frames of reference. If you did want to discuss what happens to the energy during the two stages, you would need to do the entire problem (both stages) all in one inertial reference frame. But since you seem to be more comfortable with the method of switching reference frames, it is probably best to just keep with the way you are doing it now.

edit: uh this paragraph was a reply to Tanya's last post before she edited it, which is why it seems kinda strange now.

What was your method to get the answer using the small angle approximation? You can probably do a similar method, but instead don't use small angle approximation, to get to the answer.

Last edited: Jul 21, 2014
18. Jul 21, 2014

### Tanya Sharma

Could you elaborate ?

I worked with $\frac{d^2θ}{dt^2} = -\frac{g}{L}θ$ and $\frac{d^2θ}{dt^2} = -\frac{g-a}{L}θ$ and equated the maximum speed of the bob .

How should I deal with $\frac{d^2θ}{dt^2} = -\frac{g}{L}sinθ$ and $\frac{d^2θ}{dt^2} = -\frac{g-a}{L}sinθ$ ?

19. Jul 21, 2014

### BruceW

Well, if you did everything (both stages) all in one inertial reference frame, then it is perfectly reasonable to calculate what happens to the energy as a function of time, for any time during either of the two stages.

deal with it in a similar way. equate the maximum speed of the bob. Of course, right now, the equation does not contain the angular velocity as a variable. But neither did the approximate equations. I'm guessing you did something to the approximate equations so that they would become a function of angular velocity and angular position? You can do a similar thing for the exact equations.

20. Jul 21, 2014

### ehild

Perhaps you read the end of Post #13?

ehild