# Physical Pendulum: Rotational Simple Harmonic Motion.

Question: A straight uniform stick having a length b (m) and a mass M (kg) is freely pivoted at one end. Find the frequency of oscillation about the pivot, assuming the angle theta is small.

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The problem can be solved easily by considering rotational simple harmonic motion.

For a linear simple harmonic motion force acting on a particle is directly proportional to the displacement of the particle from equilibrium position and directed to the equilibrium position and hence the equation of motion is given by

F = - K*x

Where F is the force and K is the constant may be called the force constant.

The frequency of oscillation is given by

Where m is the mass of the particle

In the same way a body is called in rotational SHM for small angles if the torque acting on it is directly proportional to the angular displacement and given by

= - K*

and the frequency of small oscillation is given analogues to the above equation as

---------------------------- (1)

Where I is the moment of inertia of a body

Now if the rod is slightly displaced from its equilibrium position, the only restoring torque acting on it is due to its weight Mg and is given by

---------------------------- (2)

Negative sign due to restoring torque

If is small enough and measured in radians we know that sin = and hence we have

---------------------------- (2)

And hence the value of K for the rotational SHM can be given by

---------------------------- (3)

The moment of the inertia of the stick about one end is given by

---------------------------- (4)

Hence substituting values from equations 4 and 3 in 1 we get the frequency of small oscillations of the rod as

© BrainMass Inc. brainmass.com December 24, 2021, 6:56 pm ad1c9bdddf>https://brainmass.com/physics/torques/physical-pendulum-rotational-simple-harmonic-motion-151553