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Question
The potential energy function for a particle executing linear SHM is given by `V(x) = 1/2 kx^2` where k is the force constant of the oscillator (Figure). For k = 0.5 N/m, the graph of V(x) versus x is shown in the figure. A particle of total energy E turns back when it reaches `x = ±x_m`. If V and K indicate the P.E. and K.E., respectively of the particle at `x = +x_m`, then which of the following is correct?
Options
V = O, K = E
V = E, K = O
V < E, K = O
V = O, K < E
Solution
V = E, K = O
Explanation:
`U = 1/2 kx^2`
For extreme positions,
`U = 1/2 ka^2`
Total energy, E = `1/2 ka^2`
`K = 1/2 mv^2` = 0
At any position, K.E., `K = E - U = 1/2 ka^2 - 1/2 kx^2`
`K = 1/2 k(a^2 - x^2)`
Total energy,
E = PE + 0 = PE
⇒ `V(x_m) = 1/2 kx_m^2`
As a result, V = E, K = O
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