CameraIcon

SearchIcon

MyQuestionIcon

235

You visited us 235 times! Enjoying our articles? Unlock Full Access!

COE in SHM

A particle is...

Question

A particle is executing linear simple harmonic motion about the origin x = 0. Which of the graphs shown in the figure represents the variation of the potential energy function U(x) versus displacement x?

A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D

Here, $F = - k x \Rightarrow \frac{- d U}{d x} = - k x \Rightarrow \int_{0}^{U} d U = \int_{0}^{x} k x d x ∴ U = \frac{1}{2} k x^{2}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. For a particle executing simple harmonic motion, the displacement x is given by

x = A c o s ω t

. Identify the graph, which represents the variation of potential energy (U) as a function of time t and displacement x.

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 displacement - time

(x - t)

graph of a particle executing simple harmonic motion is shown in figure. The correct variation of net force

F

acting on the particle as a function of time is

Q. The plot of velocity (v) versus displacement (x) of a particle executing simple harmonic motion is shown in figure. The time period of oscillation of particle is

Q. The potential energy function for a particle executing linear simple harmonic motion is given by

V (x) = k x^{2} / 2

, where k is the force constant of the oscillator.

k = 0.5 N m^{- 1}

, the graph of V(x) versus x is shown in Fig. Show that a particle of total energy 1 J moving under this potential must turn back when it reaches

x = \pm 2 m

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