Show that the angular wave number k for a nonrelativistic free particle of mass m can be written as
`k=(2pisqrt(2mk))/(h)`
in which K is the particle's kinetic energy.

The relation between wavelength (lamda) and angular wave number (k) of a wave is

Kinetic energy of a particle moving in a straight line varies with time t as K = 4t^(2) . The force acting on the particle

A particle of mass m has half the kinetic energy of another particle of mass m/2 . If the speed of the heavier particle is increased by 2 ms^(-1) its new kinetic energy becomes equal to t he original kinetic energy of the lighter particle. The ratio of the orighinal speeds of the lighter and heavier particle is

Kinetic energy of the particle performing S.H.M. is

A particle is exeuting SHM . At a point x = A//3 , kinetic energy of the particle of the particle is K , where A is the particle At a point x = 2A//3 , kinetic energy of the particle will be:

The deBroglie wavelength of a particle of kinetic energy K is lamda . What would be the wavelength of the particle, if its kinetic energy were (K)/(4) ?

The kinetic energy K of a particle moving along a circle of radius R depends upon the distance s as K=as^2 . The force acting on the particle is

Kinetic energy of a particle performing S.H.M.

If L, M and P are the angular momentum, mass and linear momentum of a particle respectively, which of the following represents the kinetic energy of the particle when the particle rotates in a circle of radius R?