Work, Power and Energy
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Question : 4
Total: 30
The potential energy function for a particle executing linear simple harmonic motion is given by V ( x ) =
k k = 0.5 N m – 1 V ( x ) x x = ± 2
Solution:
The total energy of an oscillator is the sum of kinetic energy and potential energy at any instant.
E =
m u 2 +
k x 2
The particle turn back at the instant, when its velocity becomes zero,u = 0
ThereforeE =
m u 2 +
k x 2 = 0 +
k x 2
∴1 =
×
x 2 ( ∵ E = 1 J , k =
N ∕ m )
1 =
x 2
x 2 = 4 , x = ± 2
Thus, the particle of total energy 1 J moving under this potential, must turn back atx = ± 2
Total energy = Kinetic energy + Potential energy
The particle turn back at the instant, when its velocity becomes zero,
Therefore
∴
Thus, the particle of total energy 1 J moving under this potential, must turn back at
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