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At what distance from the mean position would the K.E of a particle in simple harmonic motion be equal to its potential energy?
- a)a/√2
- b)a/2
- c)2√a
- d)a
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
At what distance from the mean position would the K.E of a particle in...
Let say from some distance x, the KE = PE and as total energy must be conserved and TE = -½ kA2
Thus we get 2PE = ½ kA2
Thus we get 2kx2 = kA2
We get x = A / √2
Thus we get 2PE = ½ kA2
Thus we get 2kx2 = kA2
We get x = A / √2
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At what distance from the mean position would the K.E of a particle in...
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At what distance from the mean position would the K.E of a particle in...
In simple harmonic motion, the relationship between kinetic energy (K.E) and potential energy (P.E) can be expressed as:
K.E = (1/2) k x^2
P.E = (1/2) k x^2
where k is the spring constant and x is the displacement from the mean position.
To find the distance from the mean position where K.E is equal to P.E, we can set the equations equal to each other:
(1/2) k x^2 = (1/2) k x^2
Simplifying and canceling the common terms, we get:
x^2 = x^2
This equation is true for all values of x, meaning that the kinetic energy will be equal to the potential energy at any distance from the mean position.
K.E = (1/2) k x^2
P.E = (1/2) k x^2
where k is the spring constant and x is the displacement from the mean position.
To find the distance from the mean position where K.E is equal to P.E, we can set the equations equal to each other:
(1/2) k x^2 = (1/2) k x^2
Simplifying and canceling the common terms, we get:
x^2 = x^2
This equation is true for all values of x, meaning that the kinetic energy will be equal to the potential energy at any distance from the mean position.
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