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[" The potential energy function for a "],[" particle executing linear simple "],[" harticle executing liven by "V(x)=],[" harmonic motton is given by "V(x)=],[kx^(2)/2" ,where "k" is the force constant "],[" of the oscillator.For "k=0.5Nm^(-1)" ."],[" the graph of "V(x)" versus "x" ls shown "],[" the graph of "V(x)" versus "x" ls shown "],[" the graph of "V(x)" versus "x" ls shown "],[" to fyg."6.12" .Show that a particle of "],[" total energy "1J" moving under this "],[" potential must turn back' when "],[" reaches "x=+-2m.]

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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 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 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 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 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?

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?

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 ?

The energy of a particle executing simple harmonic motion is given by E=Ax^2+Bv^2 where x is the displacement from mean position x=0 and v is the velocity of the particle at x then choose the correct statement(s)