Applications of Harmonic Motion: Problems 1

Previous Next

Problem : A disk of mass 2 kg and radius .5 m is hung from a wire, then rotated a small angle such that it engages in torsional oscillation. The period of oscillation is measured at 2 seconds. Given that the moment of inertia of a disk is given by I = , find the torsional constant, κ, of the wire.

To solve this problem we use the equation for the period of a torsional oscillator:

T = 2Π

Solving for κ ,

κ =

We are given T, and must simply compute I. Given the dimensions of the disk, we can simply plug into the formula we are given for the moment of inertia: I =

= .25. Thus:

κ =

= 2.47

Problem : The disk from problem 1 is replaced with an object of unknown mass and shape, and rotated such that it engages in torsional oscillation. The period of oscillation is observed to be 4 seconds. Find the moment of inertia of the object.

To find the moment of inertia we use the same equation:

T = 2Π

Solving for I,

I =

>From last problem we know that κ =

, and we are given the period (4 seconds). Thus:

I =

= 1

In the last two problems we have established a method for determining the moment of inertia of any object.

Problem : A pendulum of length L is displaced an angle θ, and is observed to have a period of 4 seconds. The string is then cut in half, and displaced to the same angle θ. How does this effect the period of oscillation?

We turn to our equation for the period of the pendulum:

T = 2Π

Clearly if we reduce the length of the pendulum by a factor of 2 we reduce the period of the oscillation by a factor of 4.

Problem : A pendulum is commonly used to calculate the acceleration due to gravity at various points around the earth. Often areas with low acceleration indicate a cavity in the earth in the area, many times filled with petroleum. An oil prospector uses a pendulum of length 1 meter, and observes it to oscillate with a period of 2 seconds. What is the acceleration due to gravity at this point?

We use the familiar equation:

T = 2Π

Solving for g:

g	=
	=	= 9.87 m/s²

This value indicates a region of high density near the point of measurement- probably not a good place to drill for oil.

Problem : What is the angular velocity of a particle moving in uniform circular motion that has the same period as a mass of 2 kg on a spring with constant 8 N/m?

Recall from our comparison of circular and oscillatory motion that the angular velocity of a particle in circular motion corresponds with the angular frequency of a particle in oscillatory motion. We know the angular frequency of the mass- spring system:

σ =