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The potential energy function for a particle executing linear SHM is given by `V(x)= 1/2kx^2` where k is the force constant of the oscillator. For `k = 0.5 Nm^(-1)`, the graph of V(x) versus x is shown in the figure A particle of total energy E turns back when it reaches `x=pmx_m`.if V and K indicate the potential energy and kinetic energy respectively of the particle at `x= +x_m`,then which of the following is correct?

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The potential energy function for a particle executing linear SHM is given by V(x) =1/2kx^(2) wher e k is the force constant of the oscillator (figure). For k = 0.5 N/m ,the graph ofV(x) versus x is shown in the figure . A particle of total energy E turns back when it reaches x = pm x_(m) . If V and K indicate the PE and KE , respectively of the particle at x = +x_(m) , then which of the following is correct ?

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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 ^-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.,br>

The potential energy function for a particle executing linear simple harmonic motion is given by V{x) = kx^2//2 , wherek 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 =overset+- 2 m.

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