Response of a Damped System Under Harmonic Force

Before we jump to analysis of the damped systems we need to inspect and analyze the physical behavior of the un-damped systems which will be subjected to the harmonic force. The harmonic force which is acting on the system can be written in the following form:

$$F={{F}_{0}}\cos \omega t$$

This force acts on the body of a mass m of the un-damped system. The equation of motion can be written in the following form:

$$m\ddot{x}+kx={{F}_{0}}\cos \omega t$$

First we will derive the homogenous solution of the previous differential equation. Let’s examine the left hand side of the differential equation. So if we have

$$m\ddot{x}+kx=0$$

then the solution of that differential equation is:

$${{x}_{h}}(t)={{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t$$

In the assumed solution the 

$${{\omega }_{n}}=\sqrt{\frac{k}{m}}$$

is the natural frequency of the system

Now, just to be sure we need to check if the assumed homogenous solution is the solution of the differential equation. To do that we need to derive the first and the second derivation of the homogenous solution:

$$\begin{align} & {{{\dot{x}}}_{h}}(t)=-{{C}_{1}}{{\omega }_{n}}\sin {{\omega }_{n}}t+{{C}_{2}}{{\omega }_{n}}\cos {{\omega }_{n}}t, \\ & {{{\ddot{x}}}_{h}}(t)=-{{C}_{1}}\omega _{n}^{2}\cos {{\omega }_{n}}t-{{C}_{2}}\omega _{n}^{2}\sin {{\omega }_{n}}t. \\ \end{align}$$

The final step is to insert the second derivation and the homogenous solution into the differential equation.

$$\begin{align} & -m\omega _{n}^{2}\left( {{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t \right)+k\left( {{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t \right)=0/:m \\ & -\omega _{n}^{2}\left( {{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t \right)+\frac{k}{m}\left( {{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t \right)=0 \\ & \omega _{n}^{2}=\frac{k}{m} \\ & -\omega _{n}^{2}\left( {{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t \right)+\omega _{n}^{2}\left( {{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t \right)=0 \\ & 0=0. \\ \end{align}$$

The next step is to get the particular solution. Because the exciting force F(t) is harmonic that means that the particular solution is also harmonic and has the same frequency ω. Thus we assume a solution in the form:

$${{x}_{p}}(t)=X\cos \omega t.$$

In the previous equation the X is a constant that denotes the maximum amplitude of particular solution. To get the expression of the particular solution we need to substitute the assumed particular solution into main differential solution. But before that we need to derive the first and second derivate of the particular solution.

$$\begin{align} & {{{\dot{x}}}_{p}}(t)=-X\omega \sin \omega t, \\ & {{{\ddot{x}}}_{p}}\left( t \right)=-X{{\omega }^{2}}\cos \omega t. \\ \end{align}$$

Now we can implement the particular solution into main differential equation. By the substitution we get:

$$\begin{align} & -mX{{\omega }^{2}}\cos \omega t+kX\cos \omega t={{F}_{0}}\cos \omega t, \\ & X\cos \omega t\left( k-m{{\omega }^{2}} \right)={{F}_{0}}\cos \omega t/:\cos \omega t, \\ & X\left( k-m{{\omega }^{2}} \right)={{F}_{0}}/:\left( k-m{{\omega }^{2}} \right) \\ & X=\frac{{{F}_{0}}}{k-m{{\omega }^{2}}}=\frac{{{\delta }_{st}}}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}} \\ \end{align}$$

In previous equation the

$${{\delta }_{st}}=\frac{{{F}_{0}}}{k}$$

denotes the deflection of the mass under a force Fo and is sometimes called static deflection because fo is a constant static force. The total solution of the differential equation becomes:

$${{x}_{h}}(t)={{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t+\frac{{{F}_{0}}}{k-m{{\omega }^{2}}}\cos \omega t$$

The previous equation isn’t finished because the unknowns C1 and C2. To determine the values of these two unknowns we need to apply the boundary conditions.

First boundary conditions:

$$\begin{align} & x(t=0)={{x}_{0}} \\ & {{x}_{h}}(0)={{C}_{1}}\cos 0+{{C}_{2}}\sin 0+\frac{{{F}_{0}}}{k-m{{\omega }^{2}}}\cos 0, \\ & {{x}_{0}}={{C}_{1}}+\frac{{{F}_{0}}}{k-m{{\omega }^{2}}}, \\ & {{C}_{1}}={{x}_{0}}-\frac{{{F}_{0}}}{k-m{{\omega }^{2}}}, \\ \end{align}$$

Second boundary conditions:

$$\begin{align} & \dot{x}(t=0)={{{\dot{x}}}_{0}}, \\ & {{{\dot{x}}}_{h}}(0)=-{{C}_{1}}{{\omega }_{n}}\sin 0+{{C}_{2}}{{\omega }_{n}}\cos 0, \\ & {{{\dot{x}}}_{0}}={{C}_{2}}{{\omega }_{n}}, \\ & {{C}_{2}}=\frac{{{{\dot{x}}}_{0}}}{{{\omega }_{n}}}. \\ \end{align}$$

So now that we determine the values of C1 and C2 we can write the complete solution of the differential equation 3.3.

$${{x}_{h}}(t)=\left( {{x}_{0}}-\frac{{{F}_{0}}}{k-m{{\omega }^{2}}} \right)\cos {{\omega }_{n}}t+\left( \frac{{{{\dot{x}}}_{0}}}{{{\omega }_{n}}} \right)\sin {{\omega }_{n}}t+\left( \frac{{{F}_{0}}}{k-m{{\omega }^{2}}} \right)\cos \omega t$$

Now let’s examine the formula for maximum amplitude of the particular solution.

$$X=\frac{{{\delta }_{st}}}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}}$$

We can modify this formula to get the following expression.

$$\begin{align} & X=\frac{{{\delta }_{st}}}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}}/:{{\delta }_{st}} \\ & \frac{X}{{{\delta }_{st}}}=\frac{1}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}} \\ \end{align}$$

This formula is very important because the ratio of X/δ represents the ration of the dynamic to the static amplitude of motion and is called MAGNIFICATION FACTOR, AMPLIFICATION FACTOR, or simply AMPLITUDE RATIO.

Magnification factor of an undamped system

With help of previous graph, the response of the system can be identified to be of three types and these types are:

CASE 1.

$$0<\frac{\omega }{{{\omega }_{n}}}<1>

The denominator in amplification factor equation is positive and the response is given by the particular solution without a change. The harmonic response of the system is said to be in phase with the external force as shown below.

CASE 2.

$$\frac{\omega }{{{\omega }_{n}}}>1$$

The denominator is negative in equation 3.10 and the steady-state solution can be expressed as:

$${{x}_{p}}(t)=-X\cos \omega t$$

Where the amplitude of motion X is redefined to be a positive quantity as:

$$X=\frac{{{\delta }_{st}}}{{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}-1}$$

Since particular solution and the harmonic force have the opposite signs, the response is said to be 180 degrees out of the phase with the external force. As the ration between the harmonic frequency and the natural frequency approaches infinity the amplitude of the particular solution approaches to zero. So response of the system to a harmonic force of very high frequency is close to zero.

CASE 3.

$$\frac{\omega }{{{\omega }_{n}}}=1$$

The amplitude X given in by the equation

$$\frac{X}{{{\delta }_{st}}}=\frac{1}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}}$$

or

$$X=\frac{{{\delta }_{st}}}{{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}-1}$$

becomes infinite. This condition, for which the forcing frequency is equal to the natural frequency of the system is called resonance. To find the response for this condition, we use following expression:

$$x(t)={{x}_{0}}\cos {{\omega }_{n}}t+\frac{{{{\dot{x}}}_{0}}}{{{\omega }_{n}}}\sin {{\omega }_{n}}t+{{\delta }_{st}}\left[ \frac{\cos \omega t-\cos {{\omega }_{n}}t}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}} \right]$$