Oscillations 3
Here’s
another example of SHM – a simple pendulum – “simple” because it consists of a
string of zero mass (but length L)
and a bob of zero size (but mass m). It
is pulled back through some angle and then released. We want to describe its
subsequent motion. Once released the only forces acting on the bob are gravity
pulling down and the tension in the string. The components of the gravitational
force are which is balanced by the tension in the
string, and which causes the bob to swing back towards
equilibrium. This last force is shown in the diagram. Since the angle decreases as the bob swings back, the force is
negative. Now, the ball and string rotate and so produce a torque about the
pivot point, and this torque provides an angular acceleration to the bob
according to
(1.1)
After some arranging (1.1) may be written as
(1.2) |
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This is close to the equation for SHM – and if we confine the amplitude of swing to small angles we can use the (useful enough to remember relation for small angles measured in radians) to arrive at
(1.3) |
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We’re done! Equation (1.3) is the same equation we solved for the mass on a spring (with replacing x and g/L replacing k/m). The solution is then
(1.4) |
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this time with the natural frequency of the pendulum
One more generalization: If we rewrite Newton’s law without assuming a value for I we have
(1.5) |
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This is valid for any pendulum, e.g. a sack of potatoes swinging from a nail. (This type of pendulum is called a “physical pendulum”.) The length L is measured from the nail to the center of mass of the sack, and the moment of inertia needs to be determined.
Moral: If a system moves under the influence of a linear restoring force, or reasonable approximation thereof, it satisfies an equation with the form of (1.5). It’s motion will be sinusoidal with period equal to the square root of the constant multiplying the dependent variable ( here, or x in the case of a mass on a spring).