(a) What is the length of a simple pendulum that oscillates with a period of 3.0...
(a) What is the length of a simple pendulum that oscillates with a period of 2.6 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2? LE = LM = Enter a number. Im (b) What mass would you need to suspend from a spring with a force constant of 20 N/m in order for the mass-spring system to oscillate with a period of 2.6 s...
A certain simple pendulum has a period on earth of 1.72 s. What is its period on the surface of Mars,where the acceleration due to gravity is 3.71 m/s2 ?
Exercise 11: Simple Harmonic Motion 1. A spring-mass system oscillates with a frequency of 10 Hz when the mass is equal to 0.50 kg. What is the stiffness of the spring? With the same spring, what would the mass need to be to double the frequency? 2. A pendulum swings with a period of 1.50 seconds when the acceleration due to gravity is equal to 9.80 m/s? What is the length of the pendulum? How would this period change if...
A simple pendulum has a period of 1.74 s. What is its length? The acceleration due to gravity is 9.8 m/s2 . Answer in units of m.
What is the period (in s) of a simple pendulum of length 0.12 m and mass 0.31 kg near the surface of a planet where the gravitational acceleration is 11 m/s2? Enter a number with 2 digits behind the decimal point..
The period T of a simple pendulum is given by T=2πLg−−√T=2πLg where L is the length of the pendulum and g is the acceleration due to gravity. Assume that g = 9.80 m/s2 exactly, and that L, in meters, is lognormal with parameters μL = 0.8 and σ2L=0.05.σL2=0.05. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Find P(T > 3).
4) a) Calculate the period of a simple pendulum if the length of the string is g on the moon. gon Moon is 16of g on Earth. gon Earth = 9.8 m/s: 1.5 meter? b) Calculate the period of the simple pendulum if you take to Mars. g on Mars 3.7 m/s What is the ratio of the period of the above simple pendulum to the period of a simple pendulum which is 4 times longer? b) d) What happens...
When a pendulum with a period of 2.00000 s is moved to a new location from one where the acceleration due to gravity was 9.80 m/s2, its new period becomes 1.99900 s. By how much does the acceleration due to gravity differ at its new location? Incorrect: Your answer is incorrect. How is the period of the pendulum related to the length of the pendulum and the acceleration due to gravity at the location of the pendulum? What happens to...
An expression for the period of a simple pendulum with string length ℓ derived using calculus is T = 2(pi)sqrt{ ℓ /g } . Where g is the acceleration due to gravity. Use the data in the table to decide whether or not the pendulum in the experiment can be considered a simple pendulum. Explain your decision. Suppose have the ability to vary the mass m of the bob and the length f of the string. You decide to to...
What is the acceleration due to gravity on a planet where a 1.5 m long pendulum oscillates with the same period as a 0.8 kg mass vibrates on an 80 N/m spring?