InterviewSolution
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InterviewSolution
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Two masses `m_(1)` and `m_(2)` are suspeded togther by a massless spring of spring constnat `k` (Fig). When the masses are in equilibrium, `m_(1)` is removed. Frequency and amplitude of oscillation of `m_(2)` are A. `omega =sqrt((K)/(m_(1))), A (m_(2)g)/(K)`B. `omega =sqrt((K)/(m_(2))), A (m_(1)g)/(K)`C. `omega =sqrt((K)/(m_(1))), A=(m_(1)g)/(K)`D. `omega = sqrt((K)/(m_(1))), A =(m_(2)g)/(K)` |
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Answer» Correct Answer - B (i) When `m_(1)` is removed only `m_(2)` is left. Therefore, angular frequency: `omega = sqrt((k)/(m_(2)))` (ii) Let `x_(1)` be the extension when only `m_(2)` is left. Then, `kx_(1) = m_(2)g or x_(1) = (m_(2)g)/(k) …(1)` Similarly, let `x_(2)` be extension in equilibrium when both `m_(1)` and `m_(2)` are suspended. Then, `(m_(1)+m_(2)) g = kx_(2), :. x_(2) = ((m_(1)m_(2)g))/(k) ...(2)` From Eqs. (1) and (2) amplitude of oscillation, `A =x_(2) -x_(1) = (m_(1)g)/(k)` |
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