InterviewSolution
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InterviewSolution
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A cord is attached between a 0.50 kg block and a string with force constant `k=20 N//m`.The other end of the spring is attrached to the wall and the cord is placed over a pulley `(I=0.60 MR^2)` of mass 5.0 kg and radius 0.50 m.(See the accompanying figure.) Assuming no slipping occurs what is the frequency of the oscillations when the body is set into motion ? |
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Answer» Correct Answer - C In equilibrium `kx_(0) = mg` `:. x_(0) = (mg)/(k) = (0.5 xx 10)/(20)` `= 0.25m` When displaced downwards by `x` from the mean position, total mechanical energy, `E = - mgx + (1)/(2)mv^(2) + (1)/(2)Iomega^(2) + (1)/(2)(x + x_(0))^(2)` Substituting `I = 0.6 MR^(2)` and `omega = (v)/(R)` We have, `E = - mgx + (1)/(2)mv^(2) + (1)/(2) (0.6MR^(2)) ((v)/(R))^(2) + (1)/(2)k (x + x_(0))^(2)` Since E = constant `:. (dE)/(dt) = 0` or `0 = - mg ((dx)/(dt)) + (1)/(2)m (2v.(dv)/(dt)) + (0.6M) + (v)(dv)/(dt) + (1)/(2)k[2(x + x_(0))] (dx)/(dt)` putting `(dx)/(dt) = v, kx_(0) = mg` and `(dv)/(dt) = a` We have `0 = (ma + 0.6 Ma) + kx` or `a = - ((k)/(m + 0.6M))x` Since, `a prop - x` motion is SHM. `f = (1)/(2pi) sqrt|(a)/(x)|` `= (1)/(2pi)sqrt((k)/(m + 0.6M))` ` = (1)/(2pi) sqrt((20)/(0.5 xx 0.6 xx 5))` ` = 0.38 Hz` . |
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