Why do you not double the length of a pendulum in order to double the period?

AnsOther people have given formal solutions, but that doesn’t really help intuitively. Allow me to do what I can to alleviate that absence.

The logic is somewhat similar to the square in the equation  

x(t)=

1

2

a

t

2

x(t)=12at2

, the equation for motion with zero initial velocity.

In that equation, if you ask how long it’ll take for some change in position  

Δx

Δx

to occur, the answer is  

t=

2Δx

g

−

−

−

√

t=2Δxg

.

The logic is vaguely* similar. Let me give you a rough argument - it’s not perfect, but has something resembling the truth.

In order to accumulate a change in position from rest, you need to first accumulate velocity, then [so to speak] position. This results in the quadratic relationship for time for free-fall. Roughly the same logic results for a pendulum: for the same angle, you’re increasing the distance that the pendulum has to travel. This means that it needs to accumultate more velocity, then more position, compared to what it was doing before; since you increase both, that quadruples the time, not just doubles it.

If that helps you understand it, great. If not, I would still generally advice you to consider the closely related question of why free-fall works the way it does, because that’s a simpler case that might give you an understanding of what’s going on more easily.

*Using dimensional analysis, the logic is exactly the same, but that’s not what I’m doing here.wer:

Explanation:

resonant frequency