A particle is executing SHM if `u_(1) and u_(2)` are the speeds of the particle at distances `x_(1) and x_(2)` from the equilibrium position, shown that the frequency of oscillation
`f=(1)/(2pi)((u_(1)^(2)-u_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))`

A particle is executing S.H.M. If u_(1) and u_(2) are the velocitiesof the particle at distances x_(1) and x_(2) from the mean position respectively, then

A particle is executing a linear S.H.M. v_(1) " and " v_(2) are as speeds at dis"tan"ces x_(1) " and " x_(2) from the equilibrium position. What is its ampiltude of oscillation ?

A particle executing harmonic motion is having velocities v_1 and v_2 at distances is x_1 and x_2 from the equilibrium position. The amplitude of the motion is

A particle is moving in SHM in a straight line. When the distance of the particle from equilibrium position has values x_(1) and x_(2) , the corresponding values of velocities are u_(1) and u_(2) . Show that time period of vibration is T=2pi[(x_(2)^(2)-x_(1)^(2))/(u_(1)^(2)-u_(2)^(2))]^(1//2)

(a) The motion of the particle in simple harmonic motion is given by x = a sin omega t . If its speed is u , when the displacement is x_(1) and speed is v , when the displacement is x_(2) , show that the amplitude of the motion is A = [(v^(2)x_(1)^(2) - u^(2)x_(2)^(2))/(v^(2) - u^(2))]^(1//2) (b) A particle is moving with simple harmonic motion is a straight line. When the distance of the particle from the equilibrium position has the values x_(1) and x_(2) the corresponding values of velocity are u_(1) and u_(2) , show that the period is T = 2pi[(x_(2)^(2) - x_(1)^(2))/(u_(1)^(2) - u_(2)^(2))]^(1//2)

Potential energy of a particle at position x is given by U=x^(2)-5x . Which of the following is equilibrium position of the particle?

Trajectories of two projectiles are shown in figure.Let T_(1) and T_(2) be the time periods and u_(1) and u_(2) their speeds of projection.Then