Algebra word problem super hard oscillations

Question

The frequency of oscillations of a pendulum varies inversely as the square root of the length of the pendulum. If a pendulum of length 25 feet oscillates with a frequency of 100 swings per minute, what would be the frequency of a pendulum of length 16 feet?

Answer 1

you need to set up the proportions and solve for the unknown.

If the frequency of the pendulum varied directly with the square root of the length of the pendulum then

sqrt(25) / sqrt(16) = 100 / x where x is the frequency of the 16 foot pendulum in swings per minute

But it is an inverse relationship so we invert (flip) one of the sides of the above equation

sqrt(25) / sqrt(16) = x / 100 take the square roots and solve for x

( you should find that the frequency of the 6 foot pendulum is (5/4)100 swings per minute = 125 spm )

Answer 2

To reword slightly: frequency is proportional to the inverse of the square root of the length of the pendulum.

Using f for frequency, and L for the length of the pendulum:

f = k * 1/√L

where k is a constant of proportionality. A priori it's unknown, but we are also given when L = 25, f = 100.

Plugging this in for f and L:

100 = k * 1/√25

100 = k/5

k = 500

Hence we have a general formula relating frequency and length:

f = 500 * 1/√L

Thus when L = 16,

f = 500 * 1/√16 = 500/4 = 125 swings/minute.

(Note this makes physical sense too, a shorter pendulum swings faster, and indeed 125 swings/min is faster than 100)