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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)1 / √2
- b)√2
- c)1
- d)1/2
Correct answer is option 'A'. Can you explain this answer?
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A block of mass m attached to a massless spring is performing oscillat...
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A block of mass m attached to a massless spring is performing oscillat...
Given information:
- Mass of the block = m
- Amplitude of oscillation before breaking off = A
- Amplitude of oscillation after half mass breaks off = fA
Solution:
1. Conservation of Energy:
When the mass breaks off, the total energy of the system remains constant.
2. Initial energy:
Total energy initially = Potential energy + Kinetic energy
= 1/2 k A^2 + 1/2 m v^2, where k is the spring constant, A is the amplitude, and v is the velocity of the block.
3. Final energy:
After half the mass breaks off, the potential energy becomes 1/2 k fA^2 and the kinetic energy is 1/2 * (m/2) v^2.
4. Equating initial and final energies:
1/2 k A^2 + 1/2 m v^2 = 1/2 k fA^2 + 1/4 m v^2
5. Simplifying the equation:
A^2 + m/k = f^2 A^2 + m/4k
f^2 = 1 - 1/2
f = 1/√2
Therefore, the correct answer is option 'A'.
- Mass of the block = m
- Amplitude of oscillation before breaking off = A
- Amplitude of oscillation after half mass breaks off = fA
Solution:
1. Conservation of Energy:
When the mass breaks off, the total energy of the system remains constant.
2. Initial energy:
Total energy initially = Potential energy + Kinetic energy
= 1/2 k A^2 + 1/2 m v^2, where k is the spring constant, A is the amplitude, and v is the velocity of the block.
3. Final energy:
After half the mass breaks off, the potential energy becomes 1/2 k fA^2 and the kinetic energy is 1/2 * (m/2) v^2.
4. Equating initial and final energies:
1/2 k A^2 + 1/2 m v^2 = 1/2 k fA^2 + 1/4 m v^2
5. Simplifying the equation:
A^2 + m/k = f^2 A^2 + m/4k
f^2 = 1 - 1/2
f = 1/√2
Therefore, the correct answer is option 'A'.
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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 isa)1 /√2b)√2c)1d)1/2Correct answer is option 'A'. Can you explain this answer?
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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 isa)1 /√2b)√2c)1d)1/2Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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 isa)1 /√2b)√2c)1d)1/2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 isa)1 /√2b)√2c)1d)1/2Correct answer is option 'A'. Can you explain this answer?.
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