Physics II Lab w answers_Part_39

Physics II Lab w answers_Part_39 - d dt This is related to...

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89 dt d Φ . This is related to the maximum velocity v rms of the magnet. I.2. Calculation of v rms : The maximum velocity of the magnet is clearly obtained at the equilibrium point i.e. at the bottom of the swing .The velocity v rms can be easily calculated. If M is the mass of the frame and magnet, l is the distance of the center of mass from the point of suspension of frame and magnet , and ° O is the initial release angle , then by conservation of energy, we have 2 sin 2 ) cos 1 ( 2 2 max 2 1 O O l g M l Mg I ° ° ± = - = --------------------(1) Thus 2 sin 2 max O I l g M ° ± = ----------------(2) Thus the quantity I l g M is actually the natural fig 2 frequency of small oscillations .Thus if T is the time period of small oscillations , then l g M I T ² 2 = --------------------------------------------------(3) Using eq (3) and v max = R ± max (R is the radius of the arc) we get 2 sin T R 4 v O max ° ± = ------------------------------------(4) I.4. We now give a rough argument that the maximum value of the emf ³ max = ³ o is proportional to v max .
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