A simple pendulum, such as a rock hanging from a piece of string or the inside of a grandfather clock, consists of a mass (the rock) and a support (the piece of string).
When the mass is moved a small distance away from its equilibrium point (the bottom of the arc), the mass will swing back and forth in a constant amount of time called the period. One period is the amount of time required for the mass to swing all the way to the other side and then swing back to its staring point.
Note: For the purpose of this assignment, we are making the simplifying assumption that our pendulum is swinging in a perfect vacuum, (i.e., there is no air resistance that will stop the pendulum).
The period of a simple pendulum is normally expressed using the variable T and is measured in seconds. It can be calculated using the length of the pendulum, L, and the acceleration due to gravity, g, using this expression:
The gravity that holds us on the earth and makes objects fall is familiar to all of us. What may not be quite as well-known, however, is that the acceleration due to gravity is actually determined by the mass (amount of material) and the radius of our planet (Note: The radius of a planet is the distance from the center of the planet to its surface.) If we represent the mass with the variable M (measured in kilograms) and the radius by the variable r (measured in meters), the acceleration due to gravity for a planet can be found from the expression:
where G is a constant called the Universal Gravitational Constant. This constant has the value as shown here:
1. find the mass and radius of three of the nine planets in our solar system. Be sure that the masses are expressed in kilograms and the radii are expressed in meters.
2. Using your data, calculate the gravitational acceleration on each of the three planets you selected. Note: With the masses measured in kilograms and the radii in meters, the units of gravitational acceleration would turn out to be meters per squared seconds.
3. Use the gravitational accelerations that you calculated in Step 2 to find the period of a 2 meter long simple pendulum on each of the three planets. Note: The length of a simple pendulum is normally expressed in meters, so it is sufficient to replace L with the number 2 in the period expression.
When solving a quadratic equation using the quadratic formula, it is possible for the b2 - 4ac term inside the square root (the discriminant) to be negative, thus forcing us to take the square root of a negative number. The solutions to the equation will then be complex numbers (i.e., involve the imaginary unit i).
In the real world, where might these so-called imaginary numbers be used?
2. When using a formula, we often know the value of one variable to a greater degree of accuracy than we know the others. In your opinion, what affect, if any, does it make on our use of a formula if we know the value of one variable to a greater degree of accuracy than another?
(See attached file for full problem description, including attachments.)© BrainMass Inc. brainmass.com March 4, 2021, 6:30 pm ad1c9bdddf
Mass of Planets, Gravitational Accelerations, Pendulum, Real-Life Applications of Imaginary Numbers and Accuracy of Formulas are investigated. The solution is detailed and well presented. The response received a rating of "5" from the student who originally posted the question.