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?
A. V=0, K=E
B. V=E, K=0
C. V lt K, K=0
D. V=0, K lt E
A. V=0, K=E
B. V=E, K=0
C. V lt K, K=0
D. V=0, K lt E
Correct Answer – B
At any instant, the total energy of an oscillator is the sum of kinetic energy and potential energy.
Total energy, E=K + V
At `x= +x_m`, the particle turns back, therefore its velocity at this point is zero, i.e. v=0
`therefore` K=0
`therefore E=1/2kx^2 =V`