In the introductory tutorial in this series we considered a harmonic **forcing function F _{0}cos(ωt)** applied to a translational spring and mass system oscillating on a frictionless horizontal plane with one degree of freedom. This system with mass

**m**and spring constant

**k**is shown in Figure 1 below with the spring extended by distance

**x**

It should be understood that a completely undamped spring and mass oscillating system does not exist. Friction forces will in time dissipate the initial **transient response** after which the forcing function produces a **steady state response**. However, analysing the response to a forcing function of a system with no explicit damping term has merit in that it introduces important concepts, in particular the property of **resonance**.

From the dynamic* free body diagram shown in Figure 1 we obtained the following equation of motion in terms of displacement x and time t

**sum of forces produces acceleration of mass m, not static equilibrium*

We now find a function **x(t)** that is a general solution to this equation. If you are not familiar with solving second-order non-homogeneous ordinary linear differential equations using the method of undetermined coefficients refer to the maths tutorial.

From the maths tutorial recall that the **general solution** to a second-order non-homogeneous ordinary linear differential equation is:

**the general solution for the complementary homogeneous equation**

**+**

**+**

**the particular solution for the non-homogeneous equation**

#### Step 1

Thus the first step is to find the general solution **x _{h}(t)** to the

**complementary homogeneous equation**:

From a previous tutorial we know the general solution to this equation is:

#### Step 2

We now propose a **particular solution x _{p}(t)** to equation (1) of the form:

**x _{p}(t) = [A.cos(ωt) + B.sin(ωt)]**

where **A** and **B** are **undetermined coefficients** and **ω** is the **frequency of the harmonic forcing function**.

#### Step 3

The general solution to equation (1) is **x _{h}(t) +x_{p}(t)** giving:

Equation (3) is a valid solution for all values of **ω** except **ω = ω _{n}** which gives a duplicate solution containing the function

**cos(ω**. The proposed particular solution for this condition takes the form:

_{n}t)**x _{p}(t) = [A.t.cos(ω_{n}t) + B.t.sin(ω_{n}t)]** which we develop later.

Meanwhile continuing with a particular solution to equation (1) for **ω ≠ ω _{n}** we find constants C

_{1}and C

_{2}for initial conditions x

_{0}and v

_{0}at time t = 0 where v

_{0}= dx(t)/dt.

Thus from equation (3) the **particular solution for equation (1)** for initial conditions **x _{0}** and

**v**is:

_{0}### Resulting motion

We now explore characteristics of the forced vibratory motion defined by equation (4).

To simplify the analysis:

Recall that **k = m.ω _{n}^{2}**

The first term in equation (5) with **cos(ω _{n}t)** derived from the general solution expresses the

**transient response**attributable to the free vibration of the system. In practice this response dissipates because of friction.

The second term in equation (5) with **cos(ωt)** derived from the particular solution expresses the **steady state response**.

#### Steady state response

Using the second term in equation (5) for steady state response giving:

we consider steady state response for three conditions:

- ω > ω
_{n} - ω < ω
_{n} - ω = ω
_{n}

where **ω** is the frequency of the harmonic forcing function and **ω _{n}** is the natural frequency of the free vibrating system.

We introduce a useful parameter of the forcing function called the **static amplitude δ = (F _{0} / k)** expressing the static displacement of mass

**m**under force

**F**.

_{0}If **X** is the amplitude of the harmonic steady state response, the factor **X/δ** called the **amplification factor** is a measure of the amplification effected by the forcing function.

**1. Consider ω > ω _{n}**

Figure 2 below plots the steady state response of the system when the frequency of the forcing function ω > ω_{n}. Note the response is out of phase with the forcing function by 180°.

In this example frequency ratio **ω/ω _{n}** = 1.26. Amplification factor

**(X/δ) ≈ 1.6**.

**2. Consider ω < ω _{n}**

Figure 3 below plots the steady state response of the system when the frequency of the forcing function ω < ω_{n}. Note the response is in phase with the forcing function.

In this example frequency ratio **ω/ω _{n}** = 0.73. Amplification factor

**X/δ ≈ 2.2**.

The general characteristics illustrated in Figures 2 and 3 can be deduced as follows.

Consider the term in equation (5) representing steady state response:

with respect to forcing function **F _{0}cos(ωt):**

(a) If **ω > ω _{n}** the coefficient

**X**is negative and thus out of phase with the forcing function by 180°.

(b) If **ω < ω _{n}** the coefficient

**X**is positive and thus in phase with the forcing function.

(c) The smaller the difference between **ω** and **ω _{n}** the greater the amplification factor

**X/δ**.

Figure 4 below illustrates characteristic (c). Amplification factor **X/δ** ≈ 10 when frequency ratio **ω/ω _{n}** = 0.95

**3. Consider ω = ω _{n}**

Figure 5 below shows the relationship between the frequency ratio **ω/ω _{n}** and the absolute value of amplification factor

**X/δ**.

Note the different characteristics as **ω** approaches **ω _{n}** for

**ω < ω**and

_{n}**ω > ω**.

_{n}This condition when **ω** is very close to or equals **ω _{n}** is known as

**resonance**where the theoretical amplification factor tends to infinity. In practice physical constraints limit the amplitude of vibration.

#### Combined transient and steady state response

Figures 2 to 5 illustrate the steady state condition of a free vibrating system subjected to a harmonic forcing function after transient effects are eliminated by the small amount of damping inherent in the system. We now consider the response including the transient term in **cos(ω _{n}t**) from equation (5).

Figures 6 and 7 below are plots for equation (7) with parameters **ω = 40 rad/s** and **ω _{n} = 30 rad/s** (frequency ratio = 0.75) giving the equation:

Figure 6 plots the waveform for the **sin[(ω + ω _{n})/2]** component of equation 8.

Figure 7 plots the **full waveform** of equation 8 with the waveform for the **sin[(ω - ω _{n})/2]** component superimposed.

These waveforms illustrate the phenomenon of **beating** where the amplitude varies periodically in a regular pattern. In this example the frequency of the resultant waveform is a function of **sin(35t)** and the envelope of the beat is a function of **sin(5t)**. Beating can often be detected in sound waves generated by vibration.

#### General solution to equation for condition ω = ω_{n}

For the final part of this tutorial we develop a general solution to the equation:

for the condition where ω = ω_{n}.

Recall that the general solution to non-homogeneous equation (1) comprises two parts. The first part is the general solution to the complementary homogeneous equation which in a previous tutorial was found to be:

The second part of the general solution is a particular solution to the non-homogeneous equation.

Recall from above that the particular solution **x _{p}(t) = [A.cos(ωt) + B.sin(ωt)]** for conditions where

ω > ω

_{n}and ω < ω

_{n}was not valid for the condition ω = ω

_{n}.

The proposed particular solution for the condition ω = ω_{n} takes the form*:

**x _{p}(t) = [A.t.cos(ω_{n}t) + B.t.sin(ω_{n}t)] --------- (9)**

** by the reduction of order method referred to the maths tutorial*

To find undetermined coefficients **A** and **B** in equation (11) let initial condition **t = 0**

gives: **2m.ω _{n}B - 0 = F_{0}**

Thus: **B = F _{0}/2m.ω_{n}**

Substitute in equation (11) for **B**

Gives **F _{0}.cos(ω_{n}t) - 2m.ω_{n}A = F_{0}.cos(ω_{n}t)**

Thus **A = 0**

Substituting for **A** and **B** in equation (9) gives the particular solution for condition **ω = ω _{n}** as:

Thus the general solution to equation (1) for condition **ω = ω _{n}** is:

From equation (12) find constants **C _{1}** and

**C**for initial conditions

_{2}**x**and

_{0}**v**at time

_{0}**t = 0**

where

**v**=

_{0}**dx(t)/dt**.

Substituting in equation (12) gives:

**x _{0} = C_{1} + 0 + 0**

Thus **C _{1} = x_{0}**

From equation (12):

Substituting for **t = 0** in equation (13) gives:

**v _{0} = ω_{n}C_{2}**

Thus **C _{2} = v_{0}/ω_{n}**

Giving the particular solution for equation (1) when **ω = ω _{n}** with initial conditions

**x**and

_{0}**v**:

_{0}To simplify the analysis of the motion represented by equation (14) let **x _{0}** and

**v**= 0 giving:

_{0}Figure 8 below plots the function x(t) given in equation (14).

This plot shows the harmonic resonant response (yellow) induced by forcing function (red) **F _{0}cos(ωt)** for

**ω = ω**implied by the steady state response for

_{n}**ω → ω**illustrated in Figure 5 above.

_{n}Note that line (green) **x(t) = (F _{0} / 2mω_{n}).t** expresses the increase in amplitude of the vibration. Note also the harmonic response

**sin(ω**and forcing function

_{n}t)**cos(ω**are out of phase by 90°.

_{n}t)Next: Forced vibrations with damping

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