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The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude = a) is
- a)a/2
- b)a√2
- c)a/√2
- d)a√2/3
Correct answer is option 'C'. Can you explain this answer?
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The kinetic energy and potential energy of a particle executing simple...
A/√2c)a/√3d)a/2√2
The correct answer is d) a/2√2.
The total mechanical energy of a particle executing simple harmonic motion is given by the sum of its kinetic energy (KE) and potential energy (PE), which can be expressed as:
E = KE + PE
At the maximum displacement (amplitude) of the particle (i.e. at x = ±a), all of its energy is in the form of potential energy, and its kinetic energy is zero. At the equilibrium position (x = 0), all of its energy is in the form of kinetic energy, and its potential energy is zero.
Therefore, the kinetic energy and potential energy of the particle will be equal when it is at a displacement x such that:
KE = PE
1/2mv^2 = 1/2kx^2
where m is the mass of the particle, v is its velocity, k is the spring constant, and x is the displacement from the equilibrium position.
Solving for x, we get:
x = √(mv^2/k)
At the maximum displacement (x = ±a), the velocity of the particle is zero, so we can substitute v = 0 and solve for a:
a = √(2E/k)
At the displacement where KE = PE, the particle has half of its total energy in the form of kinetic energy and half in the form of potential energy. Therefore, we can substitute E/2 for both KE and PE in the equation above and solve for x:
x = √((m/2)(v^2/k)) = √(E/2k)
Substituting the expression for a in terms of E and k, we get:
x = √(E/2k) = √((1/2)(ka^2)/2k) = √(a^2/8) = a/2√2
Therefore, the correct answer is d) a/2√2.
The correct answer is d) a/2√2.
The total mechanical energy of a particle executing simple harmonic motion is given by the sum of its kinetic energy (KE) and potential energy (PE), which can be expressed as:
E = KE + PE
At the maximum displacement (amplitude) of the particle (i.e. at x = ±a), all of its energy is in the form of potential energy, and its kinetic energy is zero. At the equilibrium position (x = 0), all of its energy is in the form of kinetic energy, and its potential energy is zero.
Therefore, the kinetic energy and potential energy of the particle will be equal when it is at a displacement x such that:
KE = PE
1/2mv^2 = 1/2kx^2
where m is the mass of the particle, v is its velocity, k is the spring constant, and x is the displacement from the equilibrium position.
Solving for x, we get:
x = √(mv^2/k)
At the maximum displacement (x = ±a), the velocity of the particle is zero, so we can substitute v = 0 and solve for a:
a = √(2E/k)
At the displacement where KE = PE, the particle has half of its total energy in the form of kinetic energy and half in the form of potential energy. Therefore, we can substitute E/2 for both KE and PE in the equation above and solve for x:
x = √((m/2)(v^2/k)) = √(E/2k)
Substituting the expression for a in terms of E and k, we get:
x = √(E/2k) = √((1/2)(ka^2)/2k) = √(a^2/8) = a/2√2
Therefore, the correct answer is d) a/2√2.
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Community Answer
The kinetic energy and potential energy of a particle executing simple...
K.E of SHM = 1/2mw²(a²-x²)
P.E of SHM = 1/2mw²x²
according to given conditions;
1/2mw²x² = 1/2mw²(a²-x²)
so, x²= a²- x²
2x² = a²
x² = a²/2
therefore,
x = a/2½
so answer option C
P.E of SHM = 1/2mw²x²
according to given conditions;
1/2mw²x² = 1/2mw²(a²-x²)
so, x²= a²- x²
2x² = a²
x² = a²/2
therefore,
x = a/2½
so answer option C
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The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude = a) isa)a/2b)a√2c)a/√2d)a√2/3Correct answer is option 'C'. Can you explain this answer?
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