The potential energy function for a particle executing linear simple harmonic motion is given by $U (_{x}) = \frac{1}{2} k x^{2}$ , where $k$ is the force constant. For $k = 0.5 N m^{- 1}$ ,the graph of $U (x)$ versus $x$ is shown in figure. Show that a particle of total energy $1 J$ moving under this potential $^{'} t u r n s b a c k^{'}$ when it reaches $x = \pm 2 m$ .

Open in App

Solution

At extreme oh SHM, KE=0
Total energy =potential energy $= \frac{1}{2} k x^{2}$
$1 = \frac{1}{2} (0.5) x^{2}$
$4 = x^{2}$
$x = \pm 2$

Suggest Corrections

Similar questions

Q. The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx²/2, 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.

Q. The potential energy function for a particle executing linear SHM is given by

V (x) = \frac{1}{2} k x^{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 the figure. A particle of total energy

E

turns back when it reaches

x = \pm x_{m}

. If

V

and

K

indicate the potential energy and kinetic energy, respectively of the particle at

x = + x_{m}

, then which of the following is correct?

The potential energy function for a particle executing linear simple harmonic motion is given by V(x) =kx²/2, 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.

Q. The potential energy function for a particle executing linear SHM is given by

\frac{1}{2} k x^{2}

where k is the force constant of the oscillator (Fig.). For k = 0.5 N/m, the graph of V(x) versus x is shown in the figure. A particle of total energy E turns back when it reaches x = x m . If V and K indicate the P.E. and K.E., respectively of the particle at x = +x m, then which of the following is correct?

The potential energy of a particle with displacement X is U(X). The motion is simple harmonic, when (K is a positive constant)

Join BYJU'S Learning Program

At extreme oh SHM, KE=0Total energy =potential energy =12kx21=12(0.5)x24=x2x=±2

At extreme oh SHM, KE=0
Total energy =potential energy $= \frac{1}{2} k x^{2}$
$1 = \frac{1}{2} (0.5) x^{2}$
$4 = x^{2}$
$x = \pm 2$