Damped Oscillations
We know that the motion of a simple pendulum, swinging in air, dies out eventually. Why does it happen ? This is because the air drag and the friction at the support oppose the motion of the pendulum and dissipate its energy gradually. The pendulum is said to execute `text(damped oscillations)`.
In damped oscillations, although the energy of the system is continuously dissipated, the oscillations remain apparently periodic.
The dissipating forces are generally the frictional forces. To understand the effect of such external forces on the motion of an oscillator, let us consider a system as shown in Fig.
Here a block of mass m oscillates vertically on a spring with spring constant k.
The block is connected to a vane through a rod (the vane and the rod are considered to be massless). The vane is submerged in a liquid. As the block oscillates up and down, the vane also moves along with it in the liquid. The up and down motion of the vane displaces the liquid, which in turn, exerts an inhibiting drag force (viscous drag) on it and thus on the entire oscillating system. With time, the mechanical energy of the block-spring system decreases, as energy is transferred to the thermal energy of the liquid and vane.
Let the damping force exerted by the liquid on the system be `F_d`. Its magnitude is proportional to the velocity v of the vane or the block.
The force acts in a direction opposite to the direction of v. This assumption is valid only when the vane moves slowly. Then for the motion along the x-axis
`F_d = �b v`
where b is a damping constant that depends on the characteristics of the liquid and the vane.
The negative sign makes it clear that the force is opposite to the velocity at every moment.
If the mass is pulled down or pushed up a little, the restoring force on the block due to the spring is `F_S = �kx`, where x is the displacement of the mass from its equilibrium position.
Total force `=F = �k x �b v`
By Newton�s second law of motion
`m a(t) = �k x(t) � b v(t)`
`m(d^2x)/(dt^2)+b(dx)/(dt)+kx=0`
The solution of this eq. describes the motion of the block under the influence of a damping force which is proportional to velocity.
The solution is found to be of the form
`x(t) = A e^(�b t//2m) cos (ω′t + phi )`
where a is the amplitude and `ω ′` is the angular frequency of the damped oscillator given by
`omega^'=sqrt((k/m)-(b^2)/(4m^2))`
If the oscillator is damped, the mechanical energy is not constant but decreases with time. If the damping is small, we can find E (t) by
`E(t)=1/2kA^2e^(-bt//m)`
Small damping means that the dimensionless ratio `(b/sqrt(km)) is much less than 1.
In damped oscillations, although the energy of the system is continuously dissipated, the oscillations remain apparently periodic.
The dissipating forces are generally the frictional forces. To understand the effect of such external forces on the motion of an oscillator, let us consider a system as shown in Fig.
Here a block of mass m oscillates vertically on a spring with spring constant k.
The block is connected to a vane through a rod (the vane and the rod are considered to be massless). The vane is submerged in a liquid. As the block oscillates up and down, the vane also moves along with it in the liquid. The up and down motion of the vane displaces the liquid, which in turn, exerts an inhibiting drag force (viscous drag) on it and thus on the entire oscillating system. With time, the mechanical energy of the block-spring system decreases, as energy is transferred to the thermal energy of the liquid and vane.
Let the damping force exerted by the liquid on the system be `F_d`. Its magnitude is proportional to the velocity v of the vane or the block.
The force acts in a direction opposite to the direction of v. This assumption is valid only when the vane moves slowly. Then for the motion along the x-axis
`F_d = �b v`
where b is a damping constant that depends on the characteristics of the liquid and the vane.
The negative sign makes it clear that the force is opposite to the velocity at every moment.
If the mass is pulled down or pushed up a little, the restoring force on the block due to the spring is `F_S = �kx`, where x is the displacement of the mass from its equilibrium position.
Total force `=F = �k x �b v`
By Newton�s second law of motion
`m a(t) = �k x(t) � b v(t)`
`m(d^2x)/(dt^2)+b(dx)/(dt)+kx=0`
The solution of this eq. describes the motion of the block under the influence of a damping force which is proportional to velocity.
The solution is found to be of the form
`x(t) = A e^(�b t//2m) cos (ω′t + phi )`
where a is the amplitude and `ω ′` is the angular frequency of the damped oscillator given by
`omega^'=sqrt((k/m)-(b^2)/(4m^2))`
If the oscillator is damped, the mechanical energy is not constant but decreases with time. If the damping is small, we can find E (t) by
`E(t)=1/2kA^2e^(-bt//m)`
Small damping means that the dimensionless ratio `(b/sqrt(km)) is much less than 1.