Lecture 3 Notes

Lecture 3 Notes - Recap of last lecture Chapter 13,...

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Recap of last lecture SHM can be described as uniform circular motion projected on a plane This gives x = A cos θ and a x = - ϖ 2 x Hence for the spring driven glider : and ϖ = √(k/m) k m T m k 1 f = = frequency and period are independent of amplitude Displacement, velocity and acceleration during SHM is given by x=Acos( ϖ t+ φ ) v x =dx/dt =- ϖ Asin( ϖ t+ φ ) a x =dv x /dt =- ϖ 2 Acos( ϖ t+ φ ) =- ϖ 2 x Chapter 13, Periodic motion
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Problem 13.28 On a horizontal frictionless table, an open-topped box, with a mass of 5.2 kg, is attached to an ideal horizontal spring which has a force constant of 375 N/m. Inside the box there is a stone with a mass of 3.44 kg, and the system oscillates with an amplitude of 7.50 cm. When the box has reached its maximum speed, the stone is suddenly plucked vertically out of the box without touching the box. Find, a) The period and b) amplitude of the resulting motion of the box. c) Without doing any calculations, is the new period greater or smaller than the original period?
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Problem 13.28 a) Since k did not change and m decreased without any calculation you know that T must decrease. New T = 0.74 sec b) The box has maximum velocity at the equilibrium point and v max = ϖ A = ∙A = 0.494 m/s (because given A =0.075m) When the stone is removed, due to conservation of momentum the box continues at the same speed it was travelling = v max .Therefore the new Amplitude A = v max (√m /√k), where m is the mass of the empty cart = 5.2kg. This gives A = 0.058 m or 5.8 cm
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Energy and momentum in SHM Mv 1 = (M+m)v 2 E 1 = ½Mv 1 2 ≠ E 2 = ½ (M+m)v 2 2 v 1 = v 2 =0 E 1 = E 2
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Lecture 3 Notes - Recap of last lecture Chapter 13,...

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