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InterviewSolution
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Frequency of a particle executing `SHM` is `10Hz`. The particle is suspended form a vertical spring. At the highest point of its oscillation the spring is unstretched. Find the maximum speed of the particle: `(g = 10 m//s^(2))` |
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Answer» Mean position of the particles is `(mg)/(K)` distance below unstretched position of spring. Therefore, amplitude of oscillation is `A = (mg)/(K) • = sqrt((K)/(m)) = 2•f = 20•` `:. (m)/(K) = (1)/(•^(2)), A = (g)/(•^(2))` Therefore, the maximum speed of particle will be `V_(max) = A• = (g)/(•^(2)) xx • =(g)/(•) = (1)/(2•)m//s` |
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