The potential energy function for a particle executing linear SHM is given by V(x) = (1/2)kx2 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 = Â± xm. If V and K indicate the potential energy and kinetic energy, respectively of the particle at x = +xm, then which of the following is correct?

Question

The potential energy function for a particle executing linear SHM is given by V(x) = (1/2)kx2 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 = Â± xm. If V and K indicate the potential energy and kinetic energy, respectively of the particle at x = +xm, then which of the following is correct? (A) V=0, K = E (B) V=E, K = 0 (C) V < E, K = 0 (D) V=0, K < E

Tardigrade Admin · Accepted Answer

At any instant, the total energy of an oscillator is the sum of kinetic energy and potential energy.
Total energy,

E = K + V

E = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2}

At

x = + x_{m}

, the particle turns back, therefore its velocity at this point is zero, i.e.

v = 0

∴ K = 0

∴ E = \frac{1}{2} k x^{2} = V

$V = 0$ , $K = E$

$V = E$ , $K = 0$

$V < E$ , $K = 0$

$V = 0$ , $K < E$

V=0, K=E

V=E, K=0

V<E, K=0

V=0, K<E

At any instant, the total energy of an oscillator is the sum of kinetic energy and potential energy. Total energy, E=K+V E=21​mv2+21​kx2 At x=+xm​, the particle turns back, therefore its velocity at this point is zero, i.e. v=0 ∴K=0 ∴E=21​kx2=V

$V = 0$ , $K = E$

$V = E$ , $K = 0$

$V < E$ , $K = 0$

$V = 0$ , $K < E$