MA1506CHAP2 - CHAPTER 2 OSCILLATIONS 2.1 THE HARMONIC...

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CHAPTER 2. OSCILLATIONS 2.1. THE HARMONIC OSCILLATOR Consider the pendulum shown. The small object, mass m , at the end of the pendulum, is moving on a circle of radius L , so the component of its velocity tangential to the circle is L ˙ θ Hence its tangential acceleration is L ¨ θ and so by * F = M * a we have mL ¨ θ = - mg sin θ. An obvious solution is θ = 0. This is called an EQUI- 1
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LIBRIUM solution, meaning that θ is a CONSTANT function. This means that if you set θ = 0 initially, then θ will remain at 0 and the pendulum will not move — which of course we know is correct. There is ANOTHER equilibrium solution, θ = π . Again, IN THEORY, if you set the pendulum EXACTLY at θ = π , then it will remain in that position for- ever. IN REALITY, of course, it won’t! Because the slightest puff of air will knock it over! So this equilibrium is very different from the one at θ = 0. This is a very important distinction! Equilibrium is said to be STABLE if a SMALL push away from equilibrium REMAINS small. If the small 2
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push tends to grow large, then the equilibrium is UNSTABLE. Obviously this is important for engi- neers! Especially you want vibrations of structures, engines, etc to remain small. Let’s look at θ = π . By Taylor’s theorem, near θ = π , we have f ( θ ) = f ( π ) + f 0 ( π )( θ - π ) + 1 2 f 00 ( π )( θ - π ) 2 + ... Now sin( π ) = 0, sin 0 ( π ) = cos( π ) = - 1, sin 00 ( π ) = - sin( π ) = 0 etc so sin( θ ) = 0 - ( θ - π ) - 0 + 1 6 ( θ - π ) 3 etc For small deviations away from π , θ - π is small, 3
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( θ - π ) 3 is much smaller, etc, so we can approximate sin( θ ) ≈ - ( θ - π ) so our equation is approximately ML ¨ θ = - mg sin θ = mg ( θ - π ) . Let φ = θ - π , so ¨ φ = ¨ θ , and now ¨ φ = g L φ. The general solution is φ = Ae ( g/L ) t + Be - ( g/L ) t so θ = φ + π = Ae ( g/L ) t + Be - ( g/L ) t + π. 4
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As you know, the exponential function grows very quickly; so even if θ is close to π initially, it won’t stay near to it very long! Very soon, θ will arrive either at θ = 0 or 2 π , far away from θ = π . The equilibrium is UNSTABLE! How long does it take for things to get out of control? That is determined by p g/L or rather p L/g , which has units of TIME. Note that it takes longer to fall over if L is large. SUMMARY : The equation ¨ φ = + g L φ is a symptom of INSTABILITY. The system is at equilibrium, but it will run away uncontrollably on a time scale fixed by p L/g . 5
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Now what about θ = 0? Here of course we use Tay- lor’s theorem around zero, f ( θ ) = f (0) + f 0 (0) θ + 1 2 f 00 (0) θ 2 + ... sin( θ ) = 0 + θ - 0 - 1 6 θ 3 + ... so sin( θ ) θ and we have approximately mL ¨ θ = - mgθ or ¨ θ = - g L θ = - ω 2 θ with ω 2 = g/L . That minus sign is crucial! General solution is C cos( ω t ) + D sin( ω t ) where C and D are arbitrary constants. Now using trigonometric identities you can show that ANY expression of the form C cos( x ) + D sin( x ) can 6
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be written as C cos( x ) + D sin( x ) = p C 2 + D 2 cos( x - γ ) where tan( γ ) = D/C . [You can see this easily by taking the scalar product of the vectors C D and cos( x ) sin( x ) .] So now we can write our general solution as θ = A cos( ωt - δ ) [ Check : this does satisfy ¨ θ = - ω 2 θ and it does con- tain TWO arbitrary constants, A and δ ].
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