# Period of pendulum in falling lift

A simple pendulum suspended from the ceiling of a stationary lift has period $T_0$. When the lift descends at steady speed, the period is $T_1$. When the lift descends with constant downward acceleration, the period is $T_2$. Explain why $T_0 = T_1 < T_2$.

I know that for small angles of oscillations, $T \approx 2\pi\sqrt\frac{l}{g}$ where $l$ is the length of the pendulum. For the first two cases, since $\sum F = 0$, the value of $g$ is apparently the same. However, I am unable to explain why the period is largest for $T_2$. If $g$ is the same for all cases, shouldn't their periods be the same?

• You should think about the fact that $g$ does not appear in the period formula because it is gravity. It is merely there because gravity is the only force acting on "normal" pendula. This is not true in an accelerating frame. – ACuriousMind Jul 6 '14 at 12:47
• Where in the question does it say the lift is falling? – Myridium Aug 4 '17 at 8:21
• @Myridium "descends" usually means falling. – Bill N Sep 7 '17 at 18:06
• @BillN - lifts descend, but they do not fall except under catastrophic failure. – Myridium Sep 11 '17 at 20:43

If you are in the elevator that is accelerating down, but you don't know that; instead you are told you are on a different planet and you have to determine the local acceleration of gravity. Then you would measure the force on a known mass and conclude $g'= F/m$ .

The situation is completely indistinguishable from the one where you are in the accelerating elevator. In other words - you are in a "world" where the restoring force on a pendulum is $mg'\sin\theta$. The derivation of the period then immediately follows along the same lines as it did for the usual situation (where $g'=g$)

The relevant parameter isn't necessarily $g$, but $g-a$, the difference between the gravitational acceleration of the pendulum's bob and the acceleration of the pendulum's pivot. If you don't want to get mired in a complicated free-body diagram, you can think of this as an application of the equivalence between an accelerating reference frame and a gravitational force.

Take the limiting case where the elevator is in free-fall (put it in orbit if you need more time to think about it). If you start the pendulum at some angle from the vertical, it will just … stay there. Since the weight and the pivot are accelerating in the same direction at the same rate, the distance between them never changes. In this case the period $T_\mathrm{freefall}$ is infinite, which is clearly longer than $T_0$.

• Where did $g - a$ come from? – Yiyuan Lee Jul 6 '14 at 12:27
• $a$ is the acceleration of the elevator. Sorry if that wasn't clear. – rob Jul 6 '14 at 12:29
• I mean, how was the expression $g - a$ derived? – Yiyuan Lee Jul 6 '14 at 12:32
• I am having great difficulties deriving $g - a$ from the equivalence principle at a high school level. Would it be alright to make the derivation more explicit? – Yiyuan Lee Jul 6 '14 at 12:46
• That is not what Lee is asking @rob. How did you derive g-a. – yolo123 Nov 23 '14 at 2:30

To answer where did the g - a come from is that form the period formula T = 2Pi * sqrt(l/g-a) when the elevator is going downwards it will be g-a but if its going upwards against the gravitational force it will be g+a.

One way to solve is by considering the non-inertial system, moving with acceleration a. In this case an inertia force must be introduced, with magnitude ma and direction opposite to acceleration. So when you write the equations of motion you have three force: gravity, tension and inertia force.

Second way, you consider inertial system, so there are still two forces but the acceleration of the pendulum ball should include the acceleration a, along the vertical. So the tension in the wire must provide this accelearation too, so it will increase.

In both cases, if you write the equations (Newton's second law) you obtained a modified equation of SHO, with a new period.