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 ` .

Explanation

Total energy of a simple harmonic oscillator ,
`E = K+V`
` 1= 1/2 mv^(2)+1/2 kx^(2) " "…(1)`
When simple 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)"]`