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A block of mass m attached to a massless spring is performing oscillatory motion of amplitude 'A' on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become fA.
The value of f is :
  • a)
  • b)
    1
  • c)
    1/2
  • d)
    √2
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A block of mass m attached to a massless spring is performing oscilla...
Potential energy of spring
Here, x = distance of block from mean position,
k = spring constant
At mean position, potential energy
At equilibrium position, half of the mass of block breaks off, so its potential energy becomes half.
Remaining energy =
Here, A' = New distance of block from mean position
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Community Answer
A block of mass m attached to a massless spring is performing oscilla...
Its simple , lengthy to explain , google it
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A block of mass m attached to a massless spring is performing oscillatory motion of amplitude 'A' on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become fA.The value of f is :a)b)1c)1/2d)√2Correct answer is option 'A'. Can you explain this answer?
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