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Two masses m_(1) and m_(2)are suspended w together by a light spring of spring constant k as shown in figure. When the system is in equilibrium, the mass m_(1)is removed without disturbing the system. As a result of this removal, mass m_(2) performs simple harmonic motion. For this situation mark the correct statement(s). |
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Answer» The amplitude of osciallation is `(m_(1)g)/k` `y_(0) =((m_(1) + m_(2))g)/k`  In equilibrium position, for resulting oscillation after removal of `m_(1)`, elongation in spring is: `y_(1) =(m_(2)g)/k` So, amplitude of oscillations is, `y_(0)-y_(1) =(m_(1)g)/k` TIME period of simple harmonic motion is, `T =2pi sqrt(m_(2)/k) rArr OMEGA = sqrt(k/m_(2))` |
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