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Question
The potential energy function for a particle executing linear simple harmonic motion is given by V(x) =`(kx^2)/2`, where k is the force constant of the oscillator. For k = 0.5 N m–1, the graph of V(x) versus x is shown in the figure. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.
Solution 1
In this scenario, the force constant k = 0.5 Nm-1 and the total energy of the particle E = 1J. The particle can reach a maximum distance of xm, where all of its total energy is converted into elastic potential energy.
`E =1/2 kx_m^2`
`=> x_m =sqrt((2E)/K) = sqrt((2xx1)/(0.5)) = sqrt4 = +- 2m`
Solution 2
Total energy of the particle, E = 1 J
Force constant, k = 0.5 N m–1
Kinetic energy of the particle, K = `1/2mv^2`
According to the conservation law:
E = V + K
`1 = 1/2 kx^2 + 1/2 mv^2`
At the moment of ‘turn back’, velocity (and hence K) becomes zero.
`:. 1 = 1/2kx^2`
`1/2xx0.5x^2 = 1`
`x^2 = 4`
`x =+-2`
Hence, the particle turns back when it reaches x = ± 2 m.
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