Question

A body attached to a spring oscillates in horizontal plane with frequency ‘n’. Its total energy is ‘E’ and the spring constant is ‘K’ then the velocity in the mean position is ______.

विकल्प

• `2 pi n sqrt ((2 E)/K)`

• `2 pi n sqrt (((2 E)/K))`

• `4 pi^2 n^2 (2 E)/K`

• `(4 pi^2)/n^2 ((2 E)/K)^(1//2)`

Answer 1

A body attached to a spring oscillates in horizontal plane with frequency ‘n’. Its total energy is ‘E’ and the spring constant is ‘K’ then the velocity in the mean position is `bbunderline(2 pi n sqrt ((2 E)/K))`.

Explanation:

Velocity at mean position,

v = ωA    ...(i)

Also, the energy of a particle performing S.H.M. at the mean position is given as,

E = `1/2 m omega^2 A^2`

∴ A = `sqrt ((2 E)/K)`

= `sqrt ((2 E)/K)    ...(because omega^2 = K/m)`

Since, ω = 2πn

Equation (i) can be written as,

v = `2 pi n sqrt ((2 E)/K)`