CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

COE in SHM

For k=0.5 N m...

Question

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.

Open in App

Solution

Total energy of the particle, E = 1 J

Force constant, k = 0.5 N m^–1

Kinetic energy of the particle, K =

According to the conservation law:

E = V + K

At the moment of ‘turn back’, velocity (and hence K) becomes zero.

∴

Hence, the particle turns back when it reaches x = ± 2 m.

Suggest Corrections

thumbs-up

0

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 simple harmonic motion is given by

U (_{x}) = \frac{1}{2} k x^{2}

k

is the force constant. For

k = 0.5 N m^{- 1}

U (x)

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

.

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

V (x) = \frac{1}{2} k x^{2}

k

is the force constant of the oscillator. For

k = 0.5 N m^{- 1}

V (x)

x

is shown in the figure. A particle of total energy

E

turns back when it reaches

x = \pm x_{m}

V

K

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

x = + x_{m}

, then which of the following is correct?

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?

Q. A potential energy function is shown in graph. A particle moving in this potential field in positive x-direction possess 3 J of kinetic energy at x = 1 m. Where will it reverse its direction?

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Energy in SHM

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon