Question

A "seconds pendulum" is one that moves through its equilibrium position once each second. (The period of the pendulum is precisely 2s. The length of a seconds pendulum is 0.992 7 m at Tokyo, Japan, and 0.994 2 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations? A simple pendulum makes 120 complete oscillations in 3.00 min at a location where g = 9.80 m/s^2. Find (a) the period of the pendulum and (b) its length. A particle of mass mass m slides without friction inside a hemispherical bowl of radius R. Show that if the particle starts from rest with a small displacement from equilibrium, it moves in simple harmonic motion with an angular frequency equal to that of a simple pendulum of length R. That is, omega V g/R. A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.450 Hz. The pendulum has a mass of 2.20 kg, and the pivot is located 0.350 m from the center of mass. Determine the moment of inertia of the pendulum about the pivot point.

Answer 1

ven 町10 simple-Tendulu ok 희 一9923--。.99849 - en is

hoof osci uations- 120 in +ime C) t = 3x60 second κά -f.no.of ose泓ffen Cime Penio T= 1-5 seeondg

a. g ス ! 0-3令me 20.87 m

displacement 2 the Pendwhm os ellers um.den the i.Amme ife in iu 2 9-8 se mg人