The potential energy of a particle executing SHM is given by V ( x )=(1/2) kx 2 where k is force constant. For k =0.5 N / m the graph of PE as a function of x is shown below. Find the position of particle having total energy 1 J moving in this potential at which it must turn back to its original position.

Question

The potential energy of a particle executing SHM is given by V ( x )=(1/2) kx 2 where k is force constant. For k =0.5 N / m the graph of PE as a function of x is shown below. Find the position of particle having total energy 1 J moving in this potential at which it must turn back to its original position. (A) ± 1 m (B) 0.1 m (C) ± 2 m (D) ± 0.2 m

Tardigrade Admin · Accepted Answer

Particle turns back when KE =0 and PE = TE =1 J (1/2) kx 2=(1/2) × 0.5 x 2=1 x2=4 or x=± 2 m

$\pm 1 m$

$0.1 m$

$\pm 2 m$

$\pm 0.2 m$

Particle turns back when
$K E = 0$ and $PE = TE = 1 J$
$\frac{1}{2} k x^{2} = \frac{1}{2} \times 0.5 x^{2} = 1$
$x^{2} = 4$ or $x = \pm 2 m$

Q. The potential energy of a particle executing SHM is given by V(x)=21​kx2 where k is force constant. For k=0.5N/m the graph of PE as a function of x is shown below. Find the position of particle having total energy 1J moving in this potential at which it must turn back to its original position.

±1m

0.1m

±2m

±0.2m

Particle turns back when KE=0 and PE=TE=1J 21​kx2=21​×0.5x2=1 x2=4 or x=±2m

$\pm 1 m$

$0.1 m$

$\pm 2 m$

$\pm 0.2 m$

Particle turns back when
$K E = 0$ and $PE = TE = 1 J$
$\frac{1}{2} k x^{2} = \frac{1}{2} \times 0.5 x^{2} = 1$
$x^{2} = 4$ or $x = \pm 2 m$