Consider a place where the gravity is one-fourth the gravity on Earth (g' = g/4), then the frequency of oscillation of a simple pendulum in that place, f, as compared to its frequency on earth is: f = 2f f = f/2 O r= 4f f = f/3 f = 9f Clear selection

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter12: Oscillatory Motion
Section: Chapter Questions
Problem 29P: The angular position of a pendulum is represented by the equation = 0.032 0 cos t, where is in...
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Consider a place where the gravity is one-fourth
the gravity on Earth (g' = g/4), then the frequency
of oscillation of a simple pendulum in that place,
f', as compared to its frequency on earth is:
f = 2f
f' = f/2
O f = 4f
f = f/3
O f = 9f
Clear selection
Transcribed Image Text:Consider a place where the gravity is one-fourth the gravity on Earth (g' = g/4), then the frequency of oscillation of a simple pendulum in that place, f', as compared to its frequency on earth is: f = 2f f' = f/2 O f = 4f f = f/3 O f = 9f Clear selection
A simple pendulum consists of a 0.8-kg
bob connected to a massless
inextensible cord with a length L = 1.1
m. The bob is set into motion and its
angular displacement is given by 0(t) =
0.11cos(wt), where e is in radians and
is in seconds. Take g = 9.8 m/s^2,
determine the mechanical energy
of
this pendulum.
Transcribed Image Text:A simple pendulum consists of a 0.8-kg bob connected to a massless inextensible cord with a length L = 1.1 m. The bob is set into motion and its angular displacement is given by 0(t) = 0.11cos(wt), where e is in radians and is in seconds. Take g = 9.8 m/s^2, determine the mechanical energy of this pendulum.
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