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8. A10 kg mass stretches a spring 70 cm in equilibrium. Suppose a 2 kg mass is attached to the spring, initially displaced 25

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= 140 me 2k9 210) = - IN=mg = ka 10x 9.8 kyo.70 k 98 0.7 0.25 m xlu) = 2m/sec using spring mass system + Kx = 0 at2 202n + 14Hence TH 140 Cos Frost + Sin stot Ito r LIH - TIP - + Using Tnghometric telentity COSA COS B-SIDA SinB = cos(A+B) ( 0.25 Cos

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