Period of a Pendulum in an Accelerating Elevator | Explained & Calculated

Calculate the period of a pendulum in an accelerating elevator. Rather than simply attributing the change in perceived gravity to being in an accelerated reference frame, this solution backs up a step to explain what a person inside the elevator would actually feel. Then looks at the change in the tension holding up the pendulum bob and its affect on the restoring force of the pendulum back toward equilibrium.

This is a very common problem in high school physics. It also shows up in introductory physics courses, AP Physics C Mechanics and on the JEE.

His greatness Walter Lewin also posed this problem in a series of online physics questions he posed to the Youtube community. Problem #89

[sreenivasulutadakaluru]

I really like your videos. Thanks for these great videos.
I want to say that, Just moving upward will not give positive sign for “a”. It also needs to be speeding upward. If a lift moving upwards and slowing “a” will be negative.

[opufy]

man these are some really impressive editing techniques, the timing and smoothness to it

[DeccaGG]

A good video with an incorrect (or, at the very least, highly misleading) explanation of the tension. The y-component of the tension and gravity do NOT cancel out - the pendulum swings along a circular arc, not a horizontal line, so the radial force components are what cancel out (in the non-inertial frame of the elevator). While your statement about vertical components cancelling out becomes true in the small-angle limit since tangential and horizontal directions become identified, it is not strictly true and can be misleading since it would not produce accurate generalizations to large angles.

Even for the standard simple pendulum, the tension F_T is given by m*(g + a_o_y)/cos(theta) where I've defined a_o_y = * t + phi) - g*sin^2(theta) (for theta in (-pi/2, pi/2), and where phi is the phase constant, theta_max the angular amplitude, omega the angular frequency). Notice that this behaves properly under various special cases (e.g., theta_max = theta = 0 gives a_o_y = 0 and F_T = m*g).

You can derive this result for F_T by first getting the angular acceleration alpha from the gravitational torque m*g*L*sin(theta) = m*L^2*alpha, then setting the tangential acceleration a_tan = L*(alpha), and then getting the y-component of that vector a_tan_y = -g*sin^2(theta), then adding that to the y-component of the radial acceleration a_rad_y = a_rad * cos(theta) with a_rad = v^2 / L where |v| = * t + phi). Of course, this could be written differently if we decided to plug in omega = sqrt(g/L), btw.

In the case of the pendulum being in an elevator, we must add a term m*a to the net force in the y-direction on the pendulum bob, resulting in an increase in F_T as you said; doing this, we'll algebraically see that the solution becomes the same as if we had just replaced "g" with "(g+a)", as it must in order to match what would be predicted by an observer in the non-inertial frame of the elevator. However this increase in F_T does not happen in the way you said, nor for the reason you said.

[nayabmughal]

It's still so confusing for me why time period decrease when elevator move upward because as we move up the g value decrease and t increase according to formila

[tuffdufroggin]

Love the diagrams! Keep up the good work😍

[Midhlaj_]

Is this same for a accelerating train.i mean instead of the acceleration being upward it is to horizontal

[naayerhs]

Why would the acceleration not appear to increase when travelling at some constant velocity. The ball hits the floor faster if travelling upwards with some constant velocity.

[adrianshi]

Why would the x component of tension grow, if the acceleration is upward, the T grows but only in the Ty direction, no?