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[" is "-V(1)" ."],[" Question "4" .The p...

[" is "-V_(1)" ."],[" Question "4" .The potential energy function "],[" for a particle executing linear simple "],[" harmonic motion is given by "V(x)=(1)/(2)kx^(2)],[" where,"k" is the force constant of the "],[" oscillator.For "k=0.5N/m" ,the graph "V(x)],[" oscillator.For "k=0.5N/m" ,the graph "V(x)],[" oscillator.For "k=0.5N/m" ,the graph "V(x)],[" orsus "x" is shown in the figure given below,"],[" Show that a particle of total energy "1J],[" moving under this potential must turn back when it reaches "x=+-2m" ."]

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The potential energy function for a particle executing simple harmonic motion is given by V(x)=(1)/(2)kx^(2) , where k is the force constant of the oscillatore. For k=(1)/(2)Nm^(-1) , show that a particle of total energy 1 joule moving under this potential must turn back when it reaches x=+-2m.

The potential energy function for a particle executing simple harmonic motion is given by V(x)=(1)/(2)kx^(2) , where k is the force constant of the oscillatore. For k=(1)/(2)Nm^(-1) , show that a particle of total energy 1 joule moving under this potential must turn back when it reaches x=+-2m.

The potential energy function for a particle executing simple harmonic motion is given by V(x)=(1)/(2)kx^(2) , where k is the force constant of the oscillatore. For k=(1)/(2)Nm^(-1) , show that a particle of total energy 1 joule moving under this potential must turn back when it reaches x=+-2m.

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 , 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 Fig. 6. 12. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches x= pm 2m .

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 "Nm"^(-1) , the graph of V(x) versus x is shown in figure . Show that a particle of total energy 1 J movng under this potential must 'turn back ' when it reaches x = pm 2 m .

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.

The kinetic energy (K) and potential energy (V) of a particle performing a simple harmonic motion are such that, always