Question

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 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 = ± 2 m.

Answer 1

hope it will help.........

Answer 2

$\large\sf {Given:}$

Particle energy E = $\tt 1J$

K = $\tt 0.5 Nm^{-1}$

$\large\sf {Solution:}$

${ \underline{ \boxed{formula \to \: kinetic \: energy \: = \frac{1}{2} mv {}^{2} }}}$

Based on law of conservation of energy:

$\tt{e = v + k}$

$\to \tt {1 = \frac{1}{2} kx {}^{2} + \frac{1}{2} mv {}^{2} }$

Velocity becomes zero when it turns back

$\to \tt1 = \frac{1}{2} kx {}^{2}$

$\to \tt \: \frac{1}{2} \times 0.5 {x}^{2} = 1$

$\to \tt \: x {}^{2} = 4$

$\to \tt \: x = ± 2$

Thus, on reaching $\tt x = ±2 m$ the particle turns back