Home
Class 11
PHYSICS
A particle of mass m is executing oscill...

A particle of mass m is executing oscillations about the origin on the x-axis , Its potential energy is U (x) = k [x] = k `[x]^3` , where k is a positive constant . If the amplitude of oscillation is a, then its time period T is

A

proportional to `1sqrt(a)`

B

independent of `a`

C

proportional to `sqrt(a)`

D

proportional to `a^(3//2)`

Text Solution

Verified by Experts

`V=k|x|^(3)`
`F=(dv)/(dx)=-3k|x|^(2)`…….`(1)`
The equation of simple harmonic motion is given as
`x=a sinomega t`
`rArrm(d^(2)x)/(dt^(2))=m(-aomega^(2)sinomegat)=-momega^(2).x`……`(2)`
Using `(1)` and `(2)`, we obtain
`3k|x|^(2)=momega^(2)xrArromega=sqrt(3kx//m)`
`rArr T=2pisqrt((m)/(3kx))rArrT=2pisqrt((m)/(3kasinomegat))`
`T prop (1)/(sqrt(a))`
Promotional Banner

Topper's Solved these Questions

  • ROTATIONAL MECHANICS

    FIITJEE|Exercise Example|17 Videos

  • WORK, ENERGY & POWER

    FIITJEE|Exercise Example|18 Videos

Similar Questions

Explore conceptually related problems

A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is V(x) = k|x|^3 where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.

A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is V(x) = k|x|^3 where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.

A particle of mass m is executing oscillation about the origin on X- axis Its potential energy is V(x)=kIxI Where K is a positive constant If the amplitude oscillation is a, then its time period T is proportional

A particle of mass m is executing osciallations about the origin on the x-axis with amplitude A. its potential energy is given as U(x)=alphax^(4) , where alpha is a positive constant. The x-coordinate of mass where potential energy is one-third the kinetic energy of particle is

A particle of mass m is moving in a potential well, for which the potential energy is given by U(x) = U_(0)(1-cosax) where U_(0) and a are positive constants. Then (for the small value of x)

A particle free to move along the (x - axis) hsd potential energy given by U(x)= k[1 - exp(-x^2)] for -o o le x le + o o , where (k) is a positive constant of appropriate dimensions. Then.

A particle free to move along the (x - axis) has potential energy given by U(x)= k[1 - exp(-x^2)] for -o o le x le + o o , where (k) is a positive constant of appropriate dimensions. Then.

A particle of mass m moves in a one dimensional potential energy U(x)=-ax^2+bx^4 , where a and b are positive constant. The angular frequency of small oscillation about the minima of the potential energy is equal to

A particle free to move along the x- axis has potential energy given by U(x) = k[1 - e^(-x^(2))] for -oo le x le + oo , where k is a positive constant of appropriate dimensions. Then select the incorrect option

The potential energy of a particle of mass 'm' situated in a unidimensional potential field varies as U(x) = U_0 [1- cos((ax)/2)] , where U_0 and a are positive constant. The time period of small oscillations of the particle about the mean position-