InterviewSolution
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InterviewSolution
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Calculate the angular frequency of the system shown in fingure. Friction is absent everywhere and the threads, spring and pulleys are massless. Given that `m_(A) = m_(B) = m`. |
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Answer» Correct Answer - A::B::D Let `x_(0)` be the extension in the spring in equilibrium. Then equilibrium of `A` and `B` give, `T = kx_(0) + mg sin theta` …(i) and `2T = mg` …(ii) Here, `T` is the tension in the string. Now, suppose `A` is further displaced by `(x)/(2)` and speed of `B` at this instant will be `(v)/(2)`. Total energy of the system in this position will be, `E = (1)/(2) k(x + x_(0))^(2) + (1)/(2)m_(A)v^(2) + 1/2mB((v)/(2))^(2) + m_(A)gh_(A) - m(B)gh_(B)` or `E = (1)/(2)k(x + x_(0))^(2) + 1/2mv^(2) + 1/8mv^(2) + mgxsin theta -mg x/2` or `E = (1)/(2)k(x + x_(0))^(2) + 5/8mv^(2) + mgxsin theta - mg x/2` Since, `E` is contant, `(dE)/(dt) = 0` or `0 = k(x + x_(0))(dx)/(dt) + 5/4mv ((dv)/(dt)) + mg (sin theta)((dx)/(dt)) - (mg)/(2)((dx)/(dt))` Substituting, `(dx)/(dt) = v` rArr `(dv)/(dt) = a` and `kx_(0) + mg sin theta = (mg)/(2)` [From Eqs. (i)and(ii)] We get, `(5)/(4)m a = - kx` Since, `a prop - x` Motion is simple harmonic, time period of which is, `T = 2pi sqrt|(x)/(a)|` `= 2pi sqrt((5m)/(4k))` `:. omega = (2pi)/(T) = sqrt((4k)/(5m))` |
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