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Two simple pendulums of length 0.5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed ____ oscillations. answer given is 6/5 approx equal to 1. They have worked out saying shorter pendulum performs one oscillation more by that instant compared to longer pendulum. How is it possible? Is there any other method?
Dear student, The pendulum is modeled as a point mass at the end of a massless rod. We define the following variables: θ = angle of pendulum (0=vertical) R = length of rod T = tension in rod m = mass of pendulum g = gravitational constant We will derive the equation of motion for the pendulum using the rotational analog of Newton's second law for motion about a fixed axis, which is τ = I α where
Dear student,
We will derive the equation of motion for the pendulum using the rotational analog of Newton's second law for motion about a fixed axis, which is τ = I α where
The rotational inertia about the pivot is I = m R2. Torque can be calculated as the vector cross product of the position vector and the force. The magnitude of the torque due to gravity works out to be τ = −R m g sin θ. So we have −R m g sin θ = m R2 α
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