A particle of mass `m` executes `SHM` with amplitude 'a' and frequency 'v'. The average kinetic energy during motion from the position of equilibrium to the end is:
(1) `2pi^(2) ma^(2) v^(2)` (2) `pi^(2)ma^(2) v^(2)`
(3) `(1)/(2) ma^(2) v^(2)` (4) `4pi^(2) ma^(2) v^(2)`

A particle of mass m executes SHM with amplitude .a. and frequency .v.. The average kinetic energy during motion from the position of equilibirum to the end is :

A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilibrium to the ends is

A particle of mass 'm' executes simple harmonic motion with amplitude 'a' and frequency v. The average kinetic energy during its motion from the position of equilibrium to the end is :

A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilinrium to the end is.

A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilinrium to the end is.

A particle of mass m executes SHM with amplitude a and frequency n. What is the average kinetic energy of the particle during its motion from the position of equilibrium to the end?

Velocities of a particle executing SHM are u and v at distances x_(1)andx_(2) respectively from the mean position. Prove that the amplitude of the motion is ((v^(2)x_(1)^(2)-u^(2)x_(2)^(2))/(v^(2)-u^(2)))^(1//2)

The SI unit of energy is J=kg m^(2)s^(-2) , that of speed v is ms^(-1) and of acceleration a is ms^(-2) . What of the formulae for kinetic energy (k) given below can you rule out on the basis of dimensional arguments (m stands for the mass of the body) : (a) K = m^(2)v^(2) (b) K = (1/2)mv^(2) ("c") K= ma (d) K = (3/16)mv^(2) (e) K=(1/2)mv^(2)+ma