The period (the time required for one complete swing) of a simple pendulum varies directly as the square root of its length. If a pendulum 12 feet long has a period of 4 seconds, find the period of a pendulum 3 feet long.
12 years ago
10 years ago
4.(3)^1/2
6.92
1 year ago
To find the period of a pendulum 3 feet long, we can use the property that the period of a simple pendulum varies directly as the square root of its length.
Let's denote the period of the pendulum as T and the length of the pendulum as L. We can write this relationship as:
T β βL
Since T β βL, we can also write:
T = kβL
where k is the constant of variation.
We can use the given information to find the value of k. When the pendulum is 12 feet long, it has a period of 4 seconds. Substituting these values into the equation, we have:
4 = kβ12
Now, solve for k:
k = 4 / β12
Simplify the right side:
k = 4 / β(2 x 2 x 3)
k = 4 / (2β3)
k = 2 / β3
Now that we have the value of k, we can use it to find the period of a pendulum 3 feet long.
Substituting L = 3 into the equation T = kβL, we have:
T = (2 / β3)β3
T = 2 seconds
Therefore, the period of a pendulum 3 feet long is 2 seconds.
1 year ago
To solve this problem, we need to use the direct variation formula. The formula is as follows:
Period β βLength
Where:
Period is the time required for one complete swing.
Length is the length of the pendulum.
Using this formula, we can write the relationship between the period and length as:
Period = k * βLength
Where k is the constant of variation.
Now, let's find the value of k using the given information. We know that when the pendulum is 12 feet long, the period is 4 seconds. So we can substitute these values into the equation:
4 = k * β12
To find k, we need to isolate it. Divide both sides of the equation by β12:
4 / β12 = k
Simplifying the right side gives:
k β 1.155
Now, we have the constant of variation. We can plug this value into the formula to solve the second part of the problem.
We want to find the period of a pendulum that is 3 feet long. Using the formula, we have:
Period = 1.155 * β3
Calculating the square root of 3 gives:
Period β 1.155 * 1.732
Multiplying these values gives:
Period β 2 seconds
Therefore, the period of a pendulum 3 feet long is approximately 2 seconds.