The potential energy function for a particle executing linear simple harmonic motion is given by $V (x) = k x^{2} / 2$ , 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 Fig. Show that a particle of total energy 1 J moving under this potential must turn back when it reaches $x = \pm 2 m$ .

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Solution

Total energy of the particle, $E = 1 J$
Force constant, $k = 0.5 N / m$
Kinetic energy of the particle,
$K = (1 / 2) m v^{2}$
According to the conservation law: $E = V + K 1 = (1 / 2) k x^{2} + (1 / 2) m v^{2}$
At the moment of turn back, velocity (and hence K) becomes zero.
$1 = (1 / 2) k x^{2}$
$(1 / 2) 0.5 x^{2} = 1$
$x = 2$
Hence, the particle turns back when it reaches $x = 2 m .$

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Q. The potential energy function for a particle executing linear simple harmonic motion is given by

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versus

x

is shown in figure. Show that a particle of total energy

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^{'} t u r n s b a c k^{'}

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The potential energy function for a particle executing linear simple harmonic motion is given by V(x) =kx²/2, 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 Fig. 6.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.