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Two masses m1 and m2 are suspended together by a massless spring of force constant k. When the masses are in equilibrium, m1 is removed without disturbing the system. The amplitude of oscillation is 1. m1/g 2. m2/g 3. (m1+m2)/g 4. (m1-m2)/g?
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Two masses m1 and m2 are suspended together by a massless spring of fo...
Explanation:


  • When the masses are in equilibrium, their center of mass is at the midpoint of the spring.

  • The force acting on the spring is given by F = kx, where x is the displacement of the mass from its equilibrium position.

  • Let the equilibrium position of the masses be at x = 0.

  • When m1 is removed, the center of mass of the system shifts towards m2.

  • The new equilibrium position of the system is at x = (m2/(m1+m2)).

  • The amplitude of oscillation is equal to the displacement of the mass from its equilibrium position.

  • Hence, the amplitude of oscillation is equal to (m2/(m1+m2)).

  • Therefore, the correct option is (m2/g).


Answer: Option 2. m2/g
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Two masses m1 and m2 are suspended together by a massless spring of fo...
2?
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Two masses m1 and m2 are suspended together by a massless spring of force constant k. When the masses are in equilibrium, m1 is removed without disturbing the system. The amplitude of oscillation is 1. m1/g 2. m2/g 3. (m1+m2)/g 4. (m1-m2)/g?
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