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Question 12 In the figure, block 2 of mass 2.20 kg oscillates on the end of a spring in SHM with a period of 14.00 ms. The po
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8oin X = (1.4) Coß (ut +) da da = v=(1.47W) [- sincet +)] -3 to 3.5 x 10 s V = (1.4)/2 14X10-3 sin (22)(3.5X10-3) + + 14x10-3V3 = (5.6) m18 Now, using energy conservation ² M. + m2 V² + ₂ kx2 = Ź Kamax 2 since, >> T = 2K KE 2 xmax= 472 m (472)(2.2) (

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