Let `x_0` be the maximum compression ( or extension) in the spring. In such situation
`K = 1/2 m (0)^2 = , U = 1/2 k x_0^2`
and `E = K +U = 1/2 kx_0^2 ..... (1)`
Now, if `v_0` be the maximum speed, of the block, when at mean position, then
`K = 1/2 m (v_0)^2 = , U = 1/2 k (0) =0`
`E = K+ U = 1/2 m (v_0)^2............ (2)`
But, as the block moves from its mean position, to either of its extreme position, it utilizes, its kinetic energy `(1/2 mv_0^2)` to elongate or compress; the spring, which gets stored in the spring in the form of potential energy`(1/2 k x_0^2)`.
Thus, from (1) and (2), it is evident at the total energy of the system (i.e., spring + block) remains constant.
`text(Note :)`
Oscillation of a spring pendulum is independent of the value of `g` at the place
Horizontal oscillations, of a block attached to a spring, on a rough surface.
Consider a block of mass 'm', attached to a spring of force constant k, oscillation on a rough horizontal surface with
coefficient of friction `mu`.
Let `x_0` be the initial compression, of the spring and `x_1` be the maximum extension, for the first time. From energy consideration
`1/2 kx_0^2 =1/2 kx_1^2 =(mumg)(x_0+x_1)`
`k/2(x_0+x_1)(x_0-x_1) = mu mg(x_0+x_1) => x_0-x_1 = (2 mu mg)/k`
Let `x_0` be the maximum compression ( or extension) in the spring. In such situation
`K = 1/2 m (0)^2 = , U = 1/2 k x_0^2`
and `E = K +U = 1/2 kx_0^2 ..... (1)`
Now, if `v_0` be the maximum speed, of the block, when at mean position, then
`K = 1/2 m (v_0)^2 = , U = 1/2 k (0) =0`
`E = K+ U = 1/2 m (v_0)^2............ (2)`
But, as the block moves from its mean position, to either of its extreme position, it utilizes, its kinetic energy `(1/2 mv_0^2)` to elongate or compress; the spring, which gets stored in the spring in the form of potential energy`(1/2 k x_0^2)`.
Thus, from (1) and (2), it is evident at the total energy of the system (i.e., spring + block) remains constant.
`text(Note :)`
Oscillation of a spring pendulum is independent of the value of `g` at the place
Horizontal oscillations, of a block attached to a spring, on a rough surface.
Consider a block of mass 'm', attached to a spring of force constant k, oscillation on a rough horizontal surface with
coefficient of friction `mu`.
Let `x_0` be the initial compression, of the spring and `x_1` be the maximum extension, for the first time. From energy consideration
`1/2 kx_0^2 =1/2 kx_1^2 =(mumg)(x_0+x_1)`
`k/2(x_0+x_1)(x_0-x_1) = mu mg(x_0+x_1) => x_0-x_1 = (2 mu mg)/k`