Physics II Lab w answers_Part_39

Physics II Lab w answers_Part_39 - 89 dt d . This is...

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Unformatted text preview: 89 dt d . 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 =--------------------------------------------------(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...
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Physics II Lab w answers_Part_39 - 89 dt d . This is...

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