Undamped free vibration - exercises 2

Example 1.5

The body of arbitrary shape has a mass m, mass center at G, and the radius of gyration about G of kG. If it is displaced a slight amount from its equilibrium position and released determine the natural period of vibration.

Figure 1.1 - Body of arbitrary shape

Solution: First thing in solving this example is to draw a free body diagram. From this diagram we can derive the differential equation. On next figure we will add the external and effective forces to the system.

Figure 1.2 - FBD method applied on the body

By applying Newton’s second law we can derive the differential equation. 

$$\begin{align} & \sum{{{M}_{0}}={{I}_{0}}\alpha } \\ & -mgd\sin \theta =\left[ mk_{G}^{2}+m{{d}^{2}} \right]\ddot{\theta } \\ & \ddot{\theta }+\frac{gd}{k_{G}^{2}+{{d}^{2}}}\sin \theta =0 \\ \end{align}$$

To simplify this equation we will use the approximation:

$$\sin \theta \approx \theta $$

With this approximation we get: 

$$\ddot{\theta }+\frac{gd}{k_{G}^{2}+{{d}^{2}}}\theta =0$$

From the previous differential equation we can determine the value of natural frequency: 

$${{\omega }_{n}}=\sqrt{\frac{gd}{k_{G}^{2}+{{d}^{2}}}}$$

And the period is:

$$\tau =\frac{2\pi }{{{\omega }_{n}}}=\frac{2\pi }{\sqrt{\frac{gd}{k_{G}^{2}+{{d}^{2}}}}}=2\pi \sqrt{\frac{k_{G}^{2}+{{d}^{2}}}{gd}}$$

Example 1.6

A square plate has a mass m and is suspended at its corner from a pin O. Determine the natural period of vibration if it is displaced a small amount and released.
 

Figure 1.3 - Square Plate

The mass moment of inertia can be calculated from: 
$$\begin{align} & {{J}_{0}}=\frac{1}{12}m\left( {{a}^{2}}+{{a}^{2}} \right)+m{{\left( \frac{a\sqrt{2}}{2} \right)}^{2}} \\ & {{J}_{0}}=\frac{2m{{a}^{2}}}{12}+\frac{2m{{a}^{2}}}{4}=\frac{m{{a}^{2}}}{6}+\frac{m{{a}^{2}}}{2} \\ & {{J}_{0}}=\frac{2m{{a}^{2}}}{3} \\ \end{align}$$

Now we need to calculate the sum of the moments around the point O.

$$\begin{align} & \sum{{{M}_{0}}={{I}_{0}}\alpha } \\ & -mg\left( \frac{\sqrt{2}}{2}a \right)\theta =\left( \frac{2}{3}m{{a}^{2}} \right)\ddot{\theta } \\ & \ddot{\theta }+\left( \frac{3\sqrt{2}g}{4a} \right)\theta =0 \\ & \omega =\sqrt{\frac{3\sqrt{2}g}{4a}} \\ & T=\frac{2\pi }{\omega }=6.10\sqrt{\frac{a}{g}} \\ \end{align}$$

                                                       

Example 1.7

The church bell has a mass of 375 kg, a center of a mass at the point G and the radius of gyration about point D of 0.4 m. The tongue consists of a slender rod attached to the inside of the bell at C. If an 8 kg mass is attached to the end of the rod, determine the length l of the rod so that the bell will ring silent so that the natural period of vibration of the tongue is the same as that of the bell. For the calculation, neglect the small distance between C and D and neglect the mass of the rod.

Figure 1.4 - Church Bell

As same as before we are analyzing the body which rotates about a fixed point.

$$\begin{align} & +\sum{{{M}_{0}}={{I}_{0}}\alpha } \\ & mgd\sin \theta =-{{I}_{0}}\ddot{\theta } \\ & \ddot{\theta }+\frac{mgd}{{{I}_{0}}}\sin \theta =0, \\ \end{align}$$

For small rotation we can use following approximation.

$$\sin \theta \approx \theta $$

With this approximation we are simplifying differential equation.

$$\ddot{\theta }+\frac{mgd}{{{I}_{0}}}\sin \theta =0$$

The natural frequency of this particular dynamic system is:

$$\omega =\sqrt{\frac{mgd}{{{I}_{0}}}}$$

Time of a period can be calculated form the following equation:

$$T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{\frac{mgd}{{{I}_{0}}}}}=2\pi \sqrt{\frac{{{I}_{0}}}{mgd}}$$

In order to have equal period

$$\begin{align} & T=2\pi \sqrt{\frac{{{\left( {{I}_{O}} \right)}_{T}}}{{{m}_{T}}g{{d}_{T}}}}=2\pi \sqrt{\frac{{{\left( {{I}_{O}} \right)}_{B}}}{{{m}_{B}}g{{d}_{B}}}} \\ & \frac{{{\left( {{I}_{O}} \right)}_{T}}}{{{m}_{T}}g{{d}_{T}}}=\frac{{{\left( {{I}_{O}} \right)}_{B}}}{{{m}_{B}}g{{d}_{B}}} \\ & \frac{8\left( {{l}^{2}} \right)}{8gl}=\frac{375{{\left( 0.4 \right)}^{2}}}{375g\left( 0.35 \right)} \\ & l=0.475\text{ m} \\ \end{align}$$