The period of a pendulum is the time it takes the pendulum to swing back and forth once. If the only dimensional quantities that the period depends on are the acceleration of gravity, g, and the length of the pendulum, ℓ, what combination of g and ℓ must the period be proportional to? (Acceleration has SI units of m • s-2.)

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facundo3141592 facundo3141592 · Answer 1 · 2020-08-05T03:56:36+02:00

Answer: √(L/g)

Explanation:

Here we only work with the units:

The unit of the period is units of time, so we have:

[T] = [s]

Now, the units of the length of the pendulum are units of distance:

[L] = [m]

And the units of the acceleration are:

[g] = [m/s^2]

Now, we want to work with those two in such way that the end result is only in seconds.

First, we can see that in g we have seconds square, so we know that we should use a square root.

Then we can divide L by g in order to remove the distance unit, and to have the time unit in the numerator

[L/g] = [m*s^2/m] = [s^2]

Now we apply the square root:

[√(L/g)] = [√s^2] = [s]

Then the combination is: √(L/g)

T = k*√(L/g)

where k is the constant of proportionality.