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Two masses m1 and m2 are suspended together by a massless spring of constant K. When the masses are in equilibrium, m1 is removed without disturbing the system. The amplitude of oscillations is
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Correct option is A
With mass m2 alone, the extension of the spring l is given as m2g=kl ...(i) With mass (m1+m2), the extension l' is given by (m1+m2)g=k(l+Δl) ....(ii) The increase in extension is Δl which is the amplitude of vibration. Subtracting (i) from (ii), we get m1g=kΔl or Δl=m1gkTalk to our academic expert!
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An ideal spring with spring-constant K is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is
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