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. Two masses m1 and m2 are suspended together by a massless spring of spring constant k as shown in Fig. 2. When the masses are in equilibrium, m1 is removed without disturbing the system. Find the angular frequency and amplitude of oscillation. (5 marks)Figure 2 |
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Answer» If only mass m2 is present the extension of spring l is given by m2g=kl ⇒l= m2g/k , where k=Spring constant Now with m1 and m2 the extension be l ′ (m1 +m2 )g=kl ′ l ′ = (m1 +m2/k)g-m2g/k ⇒ A = m1g/k Now, if m2 is removed the angular frequency of oscillation ω= \(\sqrt{\frac{k}{m_2}}\) (This is same as the case of m2 attach to spring) |
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