The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = (kx^(2))/2where k is the force constantof the oscillator . For k = 0.5 "Nm"^(-1) , the graphof V(x)versusx is shown in figure . Show that a particleof 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 h

1.	The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = (kx^(2))/2where k is the force constantof the oscillator . For k = 0.5 "Nm"^(-1) , the graphof V(x)versusx is shown in figure . Show that a particleof total energy 1 J movng under this potential must 'turn back ' when it reaches x = pm 2 m.
Answer» Solution :Total ENERGY of a simple HARMONIC oscillator , `E = K+V` ` 1= 1/2 MV^(2)+1/2 kx^(2) ""…(1)` Whensimple harmonic oscillator TURN back , at this time its speed or velocity will be zero . v = 0 ` :.1 = 0 + 1/2 (0.5)x^(2) "" [ :. "From equation (1)"]`