
Consider the same structure from Worked Example #2 ([lecture 19](/course/fundamentals-of-structural-dynamics-with-python/worked-example-2)). It has a mass of $10000\:kg$ and $2\:\%$ damping. However, now the structure is subject to a harmonic force with magnitude $P_0 = 600 \:N$ and a forcing frequency $f = 0.9 \:Hz$.

<Image
	src="/images/courses/fundamentals-of-structural-dynamics-with-python/worked-example-3-part-a/img1.jpg"
	alt="Lightweight steel frame (left) and SDoF dynamic model (right). | EngineeringSkills.com"
	width="600"
	height="312"
	caption="Fig 1. Lightweight steel frame (left) and SDoF dynamic model (right)."
/>

**Using this information:**

1. Determine the dynamic magnification factor and phase shift between force and steady-state response

2. Plot the complete structural response for the time range, $0\leq t\leq 60$

3. If the harmonic force is removed at $t=11$ s, determine the mass position at $t=30$ s.

We can start by establishing some constants.

```python
# Constants
m = 10000 # (kg) Mass
xi = 0.02 # Damping ratio
P = 600 # (N) Harmonic force amplitude
f = 0.9 # (Hz) Harmonic forcing frequency
omega = 2*math.pi*f # (rads/sec) Angular harmonic forcing frequency
```

We'll use the static force and deflection from example 2 ([lecture 19](/course/fundamentals-of-structural-dynamics-with-python/worked-example-2)) to recalculate the stiffness and natural frequency of the system.

```python
P_static = 1500 # (N) Static force magnitude
Delta_static = 7 # (mm) Static displacement

# Dynamic parameters
k = P_static/(Delta_static/1000) #(N/m) Stiffness
omega_n = round(math.sqrt(k/m),3) #(rads/sec) Angular natural frequency
f_n = round(omega_n/(2*math.pi),3) # (Hz) Natural frequency
T = round(1/f_n,2) # (sec) Period of oscillation
```

### 1. DMF and phase shift, $\phi$

To determine the dynamic magnification factor we can use the following equation derived previously,

$$
DMF = \frac{1}{\sqrt{(1-\beta^2)^2 + (2\xi\beta)^2}}
\tag{1}
$$

```python
beta = omega/omega_n # Frequency ratio
DMF = 1/math.sqrt((1-beta**2)**2 + (2*xi*beta)**2) #Dynamic Magnification Factor
print(f'The frequency ratio is {round(beta,3)}.')
print(f'The dynamic magnification factor is {round(DMF,3)}.')
print(f'The static deflection is {round(P/k,4)} m.')
print(f'The DMF indicates a steady-state dynamic amplitude of {round(DMF*P/k,4)} m.')
```

- `The frequency ratio is 1.222.`
- `The dynamic magnification factor is 2.021`
- `The static deflection is 0.0028`
- `The DMF indicates a steady-state dynamic amplitude of 0.0057`

The phase shift between force and response is obtained using,

$$
\phi = \tan^{-1}\left(\frac{2\xi\beta}{1-\beta^2}\right)
\tag{2}
$$

Note that in this case since the forcing frequency is larger than the natural frequency, $\beta>1$ and we should expect a phase angle $\phi>90^{\circ}$. The denominator in the equation for $\phi$ above is negative, so we will use `arctan2` from the numpy library that will account for the negative denominator and return an angle greater than $90^{\circ}$.

```python
#Phase angle between force and response (arctan identifies correct quadrant)
Phi = np.arctan2((2*xi*beta),(1-beta**2))
print(f'The phase between force and response is {round(Phi,3)} rad.')
```

- `The phase between force and response is 3.043 rad.`

As expected, we can see that the response of the SDoF system is lagging $174.3^{\circ}$ behind the harmonic excitation force. We can further visualise this by plotting $\phi$ against $\beta$ for a range of $\beta$ values and plotting our specific case onto the graph.

```python
#Generate range of Beta values
max_B = 2 #Max value of Beta
delta_B = 0.01 #Step size in range of Beta values
nPoints = max_B/delta_B #Number of datapoints in the range
B = np.linspace(0,max_B, int(nPoints)) #Vector of Beta values

phi = np.arctan2((2*xi*B),(1-B**2))

#Set up the figure
fig = plt.figure()
axes = fig.add_axes([0.1,0.1,2,1.5])
axes.plot(B,phi*(180/pi)) #Plotting all values of phi versus beta
axes.plot([beta],[phase*(180/pi)],'ro') #Plotting our specific case with a red dot
axes.set_xlim([0,max_B])
axes.set_xlabel('Frequency ratio $\\beta$')
axes.set_ylabel('Phase Angle')
axes.set_title('Response Phase Angle')
axes.grid()
```

<Image
	src="/images/courses/fundamentals-of-structural-dynamics-with-python/worked-example-3-part-a/img2.jpg"
	alt="Variation of response phase lag with frequency ratio showing the specific phase lag in this case (red dot). | EngineeringSkills.com"
	width="1000"
	height="534"
	caption="Fig 2. Variation of response phase lag with frequency ratio showing the specific phase lag in this case (red dot)."
/>

We'll continue this solution in the next lecture where we'll plot the first $60$ seconds of the response.


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