Physics · Oscillations and Simple Harmonic Motion
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A particle executes SHM along a straight line. Its amplitude is A. The potential energy of the particle is equal to the kinetic energy when the displacement of the particle from the mean position is:
- A
Zero
- B
±A/2
- C
±A/√2
- D
2A
When the potential energy is equal to the kinetic energy, we have:
(1/2)kx^2 = (1/2)mv^2
x^2 = (mv^2)/k
v^2 = kx^2/m
Substituting v^2 into the expression for K, we have:
K = (1/2)mv^2 = (1/2)m(kx^2/m) = (1/2)kx^2
Thus, when the potential energy is equal to the kinetic energy, we have:
2K = 2U
2(1/2)kx^2 = (1/2)kA^2
x^2 = A^2/2
x = ±A/√2
Therefore, the displacement of the particle from the mean position when the potential energy is equal to the kinetic energy is ±A/√2.l
As it is numerical, it can have only one possible answer.
this option is incorrect.
When the potential energy is equal to the kinetic energy, we have:
(1/2)kx^2 = (1/2)mv^2
x^2 = (mv^2)/k
v^2 = kx^2/m
Substituting v^2 into the expression for K, we have:
K = (1/2)mv^2 = (1/2)m(kx^2/m) = (1/2)kx^2
Thus, when the potential energy is equal to the kinetic energy, we have:
2K = 2U
2(1/2)kx^2 = (1/2)kA^2
x^2 = A^2/2
x = ±A/√2
Therefore, the displacement of the particle from the mean position when the potential energy is equal to the kinetic energy is ±A/√2.l
this option is incorrect.
When the potential energy is equal to the kinetic energy, we have:
(1/2)kx^2 = (1/2)mv^2
x^2 = (mv^2)/k
v^2 = kx^2/m
Substituting v^2 into the expression for K, we have:
K = (1/2)mv^2 = (1/2)m(kx^2/m) = (1/2)kx^2
Thus, when the potential energy is equal to the kinetic energy, we have:
2K = 2U
2(1/2)kx^2 = (1/2)kA^2
x^2 = A^2/2
x = ±A/√2
Therefore, the displacement of the particle from the mean position when the potential energy is equal to the kinetic energy is ±A/√2.l
When the potential energy is equal to the kinetic energy, we have:
(1/2)kx^2 = (1/2)mv^2
x^2 = (mv^2)/k
v^2 = kx^2/m
Substituting v^2 into the expression for K, we have:
K = (1/2)mv^2 = (1/2)m(kx^2/m) = (1/2)kx^2
Thus, when the potential energy is equal to the kinetic energy, we have:
2K = 2U
2(1/2)kx^2 = (1/2)kA^2
x^2 = A^2/2
x = ±A/√2
Therefore, the displacement of the particle from the mean position when the potential energy is equal to the kinetic energy is ±A/√2.l
As it is numerical, it can have only one possible answer.
this option is incorrect.
When the potential energy is equal to the kinetic energy, we have:
(1/2)kx^2 = (1/2)mv^2
x^2 = (mv^2)/k
v^2 = kx^2/m
Substituting v^2 into the expression for K, we have:
K = (1/2)mv^2 = (1/2)m(kx^2/m) = (1/2)kx^2
Thus, when the potential energy is equal to the kinetic energy, we have:
2K = 2U
2(1/2)kx^2 = (1/2)kA^2
x^2 = A^2/2
x = ±A/√2
Therefore, the displacement of the particle from the mean position when the potential energy is equal to the kinetic energy is ±A/√2.l
Tagged under Physics · Oscillations and Simple Harmonic Motion · 2017