The Simple Pendulum An application of Simple Harmonic Motion A mass m at the end of a massless rod of length L There is a restoring force which acts to restore the mass to 0 F mg sin Compare to the spring Fs kx The pendulum does not display SHM L m T mgsin mg But for very small rad we can make the approximation 0 5 rad or about 25 simple pendulum approximation sin F mg This is SHM since s r L Arc length s mg Looks like spring F mg s force L L mg Like the spring Fs kx k constant L Now consider the angular frequency of the spring k mg L g m m L 1 g Simple pendulum f 2 L frequency Simple pendulum angular frequency With this the same equations expressing the displacement x v and a for the spring can be used for the simple pendulum as long as is small For large the SHM equations in terms of sin and cos are no longer valid more complicated functions are needed which we will not consider A pendulum does not have to be a point particle The Physical Pendulum A rigid body can also be a pendulum The simple pendulum has a moment of inertia Rewrite in terms of I 2 I mL g mg mgL 2 L mL mL mgL 1 mgL or f I 2 I L m L is the distance from the rotation axis to the center of gravity cg mg Example Use a thin disk for a simple physical pendulum with rotation axis at the rim a find its period of oscillation and b the length of an equivalent simple pendulum Solution R a From table 10 1 1 2 Icm MR M 2 But we need I at the rim so apply parallel axis theorem h R 2 1 2 2 2 3 2 Irim Icm MD MR MR MR 2 Since physical pendulum frequency is 1 f 2 MgL I T 2 I MgL Distance from rotation axis to cm L R T 2 2 MR 3R 2 MgR 2g 3 2 Let R 0 165 m 6 5 inches 3 0 165 T 2 0 999 s 2 9 8 Would make a good clock Note that the period or frequency of a pendulum does not depend on the mass and would be different on other planets b For an equivalent simple pendulum we need the simple and disk pendulums to have the same period 3R L Tdisk 2 Tsp 2 2g g 3R L 3R L 2g g 2g g 3R 3 0 165 m L 0 248 m 2 2 Damped Harmonic Motion Simple harmonic motion in which the amplitude is steadily decreased due to the action of some non conservative force s i e friction or air resistance F bv where b is the damping coefficient 3 classifications of damped harmonic motion 1 Underdamped oscillation but amplitude decreases with each cycle shocks 2 Critically damped no oscillation with smallest amount of damping 3 Overdamped no oscillation but more damping than needed for critical Apply Newton s 2nd Law The solution is Where x t A0e 2 0 b 2m 2 b t 2m cos t 0 0 k m Type of damping determined by comparing 0 and b 2m Envelope of damped motion A A0e bt 2m SHM underdamped t rad Overdamped Critically damped Underdamped SHM Forced Harmonic Motion Unlike damped harmonic motion the amplitude may increase with time Consider a swing or a pendulum and apply a force that increases with time the amplitude will increase with time Forced HM SHM t rad Consider the spring mass system it has a frequency 1 f 2 k m f0 We call this the natural frequency f0 of the system All systems car bridge pendulum etc have an f0 We can apply a time dependent external driving force with frequency fd fd f0 to the spring mass system F t Fext cos 2 f d t This is forced harmonic motion the amplitude increases But if fd f0 the amplitude can increase dramatically this is a condition called resonance Examples a out of balance tire shakes violently at certain speeds b TacomaNarrows bridge s f0 matches frequency of wind
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