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?
f = k(1/√l)
given: when l = 25 when f = 100
100 = k(1/√100) ----> k = 1000
so f = 1000(1/√l)
plug in l = 16 to find f
What is L?
To solve this problem, we can use the formula for inverse variation:
y = k/x
where y represents the frequency of oscillations, x represents the length of the pendulum, and k is a constant.
We are given that the frequency of a pendulum of length 25 feet is 100 swings per minute. Plugging this into the formula:
100 = k/25
To find the value of k, we can multiply both sides of the equation by 25:
100 * 25 = k
k = 2500
Now, we can use this value of k to find the frequency for a pendulum of length 16 feet:
y = 2500/x
y = 2500/16
Simplifying, we get:
y = 156.25
Therefore, the frequency of a pendulum of length 16 feet would be 156.25 swings per minute.
To solve this problem, we can use the concept of inverse variation. Inverse variation states that when two variables are inversely proportional, their product remains constant.
In this case, let's assume that the frequency of oscillations (f) is inversely proportional to the square root of the length of the pendulum (L). Mathematically, we can write it as:
f ∝ 1/√L
Where ∝ represents proportional.
To find the constant of variation, we need to find the product of f and √L for a given pendulum. In this case, we have a pendulum of length 25 feet oscillating with a frequency of 100 swings per minute.
So, f * √L = k, where k is the constant of variation.
Plugging in the values, we get:
100 * √25 = k
100 * 5 = k
k = 500
Now that we have the constant of variation, we can use it to find the frequency of a pendulum of length 16 feet.
Let's call this frequency f2 and length L2. We want to find f2.
Using the formula again, we have:
f2 * √L2 = k
Plugging in the values:
f2 * √16 = 500
f2 * 4 = 500
f2 = 500/4
f2 = 125
Therefore, the frequency of a pendulum of length 16 feet would be 125 swings per minute.