Step 3: The Pendulum Period Equation
With the periods of each pendulum selected, the next step is to figure out the lengths and release angles of the pendulums. In order to do this, we'll take a look at some equations.
The equation for the period of a pendulum is given by
T=2*pi*sqrt(L/g)*K(θ).
Here is a figure demonstrating L and θ. R and S are variables for the release mechanism, to be solved later.
Explanation of terms:
pi = 3.1415926..., etc.
L = the distance from the top of the string to the center of mass (the center of the pendulum bob, assuming the string is massless)
g = the acceleration due to gravity (approximately 9.81 m/s^2 or 32.17 ft/s^2)
θ = the release angle of the pendulum (the angle between the string when the pendulum is at rest and when the pendulum is released).
K(θ) = a correcting function that takes into account the effects of the release angle on the period. It is an infinite power series, but the successive terms grow very small very rapidly. A few terms suffice to get an accurate approximation.
K(θ) = 1 + θ^2/16 + 11*θ^4/3072+173*θ^6/737280+...
For angles significantly less than 1 radian, K(θ) approaches 1 and can be neglected. To test whether an angle is "significantly less" than 1 radian, take the sine of the angle in question, then compare it to the original angle. If both the angle and the sine of the angle are nearly the same, then K(θ) can likely be neglected without greatly affecting the performance of the pendulum wave.
Assuming that the release angles are small and K(θ) can be neglected, then the only variable in the equation that must be solved for is L.
As somewhat of a perfectionist, I am against neglecting K(θ), even for smaller angles. I decided to do go the harder of the two routes, suffering the lengthier calculations for (hopefully) greater accuracy and a better-performing pendulum wave.
Why do some pendulum waves that ignore K(θ) yet have large release angles still look so good?
Here's a random example I found on YouTube:
In this example, all angles are constant, and thus K(θ) is also constant. As a result, all pendulum periods are equally wrong, making all of them look right relative to each other. For my pendulum wave, K(θ) will not be constant since the release angles vary. Because of this, not taking K(θ) into account will definitely affect things.
The equation for the period of a pendulum is given by
T=2*pi*sqrt(L/g)*K(θ).
Here is a figure demonstrating L and θ. R and S are variables for the release mechanism, to be solved later.
Explanation of terms:
pi = 3.1415926..., etc.
L = the distance from the top of the string to the center of mass (the center of the pendulum bob, assuming the string is massless)
g = the acceleration due to gravity (approximately 9.81 m/s^2 or 32.17 ft/s^2)
θ = the release angle of the pendulum (the angle between the string when the pendulum is at rest and when the pendulum is released).
K(θ) = a correcting function that takes into account the effects of the release angle on the period. It is an infinite power series, but the successive terms grow very small very rapidly. A few terms suffice to get an accurate approximation.
K(θ) = 1 + θ^2/16 + 11*θ^4/3072+173*θ^6/737280+...
For angles significantly less than 1 radian, K(θ) approaches 1 and can be neglected. To test whether an angle is "significantly less" than 1 radian, take the sine of the angle in question, then compare it to the original angle. If both the angle and the sine of the angle are nearly the same, then K(θ) can likely be neglected without greatly affecting the performance of the pendulum wave.
Assuming that the release angles are small and K(θ) can be neglected, then the only variable in the equation that must be solved for is L.
As somewhat of a perfectionist, I am against neglecting K(θ), even for smaller angles. I decided to do go the harder of the two routes, suffering the lengthier calculations for (hopefully) greater accuracy and a better-performing pendulum wave.
Why do some pendulum waves that ignore K(θ) yet have large release angles still look so good?
Here's a random example I found on YouTube:
In this example, all angles are constant, and thus K(θ) is also constant. As a result, all pendulum periods are equally wrong, making all of them look right relative to each other. For my pendulum wave, K(θ) will not be constant since the release angles vary. Because of this, not taking K(θ) into account will definitely affect things.
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