or Parts I and II, use the pendulum equation and solve to predict the length needed for a pendulum with a period of 1 s and 2 s respectively.
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For Parts I and II, use the pendulum equation and solve to predict the length needed for a pendulum with a period of 1 s and 2 s respectively. Use a piece of string (or thread, dental floss, shoe lace, etc.) and a steel nut (or washer or something small but with enough mass to weigh down the string) to build each pendulum. Time each pendulum for 30 periods and then find the average time for one period. Remember, a period is the time to complete one cycle of motion, out and back. Tape each pendulum up in perhaps a doorway where it is stationary and has room to swing. Be precise with your measuring and timing. Show your work.
Part I
Given:
T = 1.00 s
l =?
Part II
Given:
T = 2.00 s
l =?
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- part A and BWhat is the period of oscillation (in seconds) of a rod of length 4.43m that is fixed at one end, but otherwise free to rotate without any friction, which has a mass 4.78kg, for small displacements? Note: In the space below, please enter you numerical answer. Do not enter any units. If you enter units, your answer will be marked as incorrect.DONT MIND THE BIG NUMBER ON THE RIGHT, IT IS JUST FOR NUMBERING. MAKE SURE IT IS CORRECT AND TYPEWRITTEN TO GET AN UPVOTE. NO UPVOTE IF IT IS HANDWRITTEN. THANK YOU
- Do number 2You will fire the spring gun 3 times from the first detent and measure the change in height of the (pendulum + ball) for each shot. Write the equation for the change in height of the first shot.Item 1 Learning Goal: To understand the application of the general harmonic equation to the kinematics of a spring oscillator. One end of a spring with spring constant k is attached to the wall. The other end is attached to a block of mass m. The block rests on a frictionless horizontal surface. The equilibrium position of the left side of the block is defined to be x = 0. The length of the relaxed spring is L. (Figure 1) The block is slowly pulled from its equilibrium position to some position init> 0 along the x axis. At time t = 0, the block is released with zero initial velocity. The goal is to determine the position of the block (t) as a function of time in terms of w and init It is known that a general solution for the displacement from equilibrium of a harmonic oscillator is x(t) = C cos (wt) + S sin (wt), where C, S, and w are constants. (Figure 2) Your task, therefore, is to determine the values of C and S in terms of w and init Figure 1 of 3 L Xinit win x = 0 Part A Using the…
- Consider two pendulums with lengths l1 and I2. Express the ratio of their lengths //l2 in terms of periods T1 and T2. When submitting your answer please keep in mind that, for example T2/T2 must be written as "T1^2/T2". Please use the "Display response" button to check you entered the answer you expect. Display responseUse the following prompt for questions 1-2. Alexandra and Sal are conducting an experiment to study pendulums. In their first experiment, Alexandra and Sal changed the length of the pendulum and measured its period. In their second experiment, the mass of the pendulum bob was varied, and period was measured. 1. Before running the experiments, Sal hypothesized that increasing the mass of the pendulum would result in an increase in the pendulum's period. Was Sal correct? A. Sal was correct because mass is the only variable that will affect the period of the pendulum. B. Sal was correct because mass and length both affect the period of the pendulum. C. Sal was incorrect because only the length of the pendulum will affect the period of the pendulum. D. Sal was incorrect because only the initial angle will affect the period of the pendulum. inate Education TM, Inc.What is the period of oscillation (in seconds) of a rod of length 4.09m that is fixed at one end, but otherwise free to rotate without any friction, which has a mass 4.18kg, for small displacements? Note: In the space below, please enter you numerical answer. Do not enter any units. If you enter units, your answer will be marked as incorrect.