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A particle is viberating in a simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position, is its energy half potential and kinetic ?
- a)3 cm
- b)2√2
- c)1 cm
- d)√2 cm
Correct answer is option 'B'. Can you explain this answer?
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A particle is viberating in a simple harmonic motion with an amplitude...
Understanding the Problem:
Given that the particle is vibrating in simple harmonic motion with an amplitude of 4 cm, we need to find the displacement from the equilibrium position at which its energy is half potential and half kinetic.
Calculating the Energy:
- Total energy of a particle in simple harmonic motion is the sum of its potential energy and kinetic energy.
- At any given point, the total energy remains constant and is equal to the maximum potential energy or kinetic energy.
- When the particle is halfway between the equilibrium position and the extreme position, its energy is equally divided between potential and kinetic energy.
Solving for Displacement:
- At a displacement x from the equilibrium position, the potential energy is given by PE = (1/2)kx², where k is the spring constant.
- The kinetic energy is given by KE = (1/2)mv², where m is the mass of the particle and v is its velocity.
- At the point where the energy is equally divided, PE = KE.
- Therefore, (1/2)kx² = (1/2)mv².
- Since x = 4 cm, we can solve for v and then find the displacement at which energy is half potential and half kinetic.
Calculating the Displacement:
- We find that v = √(k/m)x.
- Substituting the given values, v = √(k/m)*4.
- When PE = KE, (1/2)kx² = (1/2)mv² simplifies to kx² = mv².
- Substituting the values and simplifying, we get √(k/m)*16 = 4.
- Solving for x, x = 2√2 cm.
Therefore, the displacement from the equilibrium position at which the particle's energy is half potential and half kinetic is 2√2 cm, which corresponds to option 'B'.
Given that the particle is vibrating in simple harmonic motion with an amplitude of 4 cm, we need to find the displacement from the equilibrium position at which its energy is half potential and half kinetic.
Calculating the Energy:
- Total energy of a particle in simple harmonic motion is the sum of its potential energy and kinetic energy.
- At any given point, the total energy remains constant and is equal to the maximum potential energy or kinetic energy.
- When the particle is halfway between the equilibrium position and the extreme position, its energy is equally divided between potential and kinetic energy.
Solving for Displacement:
- At a displacement x from the equilibrium position, the potential energy is given by PE = (1/2)kx², where k is the spring constant.
- The kinetic energy is given by KE = (1/2)mv², where m is the mass of the particle and v is its velocity.
- At the point where the energy is equally divided, PE = KE.
- Therefore, (1/2)kx² = (1/2)mv².
- Since x = 4 cm, we can solve for v and then find the displacement at which energy is half potential and half kinetic.
Calculating the Displacement:
- We find that v = √(k/m)x.
- Substituting the given values, v = √(k/m)*4.
- When PE = KE, (1/2)kx² = (1/2)mv² simplifies to kx² = mv².
- Substituting the values and simplifying, we get √(k/m)*16 = 4.
- Solving for x, x = 2√2 cm.
Therefore, the displacement from the equilibrium position at which the particle's energy is half potential and half kinetic is 2√2 cm, which corresponds to option 'B'.
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A particle is viberating in a simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position, is its energy half potential and kinetic ?a)3 cmb)2√2c)1 cmd)√2 cmCorrect answer is option 'B'. Can you explain this answer?
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