Chapter 11, Problem 88GP

To determine

To Find:The frequency of vibration.

Answer to Problem 88GP

The frequency of vibration is $f=2.5\text{ Hz}$ .

Explanation of Solution

Given Information:

Shear modulus, $G=520{\text{N/m}}^{\text{2}}$

Density, $\rho =1300{\text{kg/m}}^{\text{3}}$

Length, $L=4\text{cm}$

Area = $32{\text{cm}}^{\text{2}}$

Concept used:

Shear strain can be given as:

  $\Delta L=\frac{FL}{GA}$

Here, $F$ is the force exerted.

  $G$ is the shear modulus

  $A$ is the area of cross-section,

  $L$ is the length,

  $\Delta L$ is the change in length.

Calculation:

Rearranging the above expression,

  $F=\frac{GA}{L}\Delta L$

Also, comparing $F=k\Delta L$ with the above expression,

  $k=\frac{GA}{L}$

Hence, the frequency can be calculated using:

  $\begin{array}{l}\omega =\sqrt{\frac{k}{m}}\\ 2\pi f=\sqrt{\frac{\left(\frac{GA}{L}\right)}{m}}=\sqrt{\frac{GA}{mL}}\\ f=\frac{1}{2\pi }\sqrt{\frac{GA}{\left(\rho V\right)L}}\\ f=\frac{1}{2\pi }\sqrt{\frac{GA}{\left(\rho \left(AL\right)\right)L}}\\ f=\frac{1}{2\pi }\sqrt{\frac{G}{\rho L^2}}\end{array}$

Substituting all the values:

  $\begin{array}{l}f=\frac{1}{2\left(3.14\right)}\sqrt{\frac{\left(520{\text{ N/m}}^2\right)}{\left(1300{\text{ kg/m}}^3\right){\left(4\times {10}^{-2}\text{m}\right)}^2}}\\ f=2.5\text{ Hz}\end{array}$

Conclusion:

The frequency of vibration is $f=2.5\text{ Hz}$ .