Chapter 11, Problem 20P

To determine

To Calculate:The frequency of vibration.

Answer to Problem 20P

The frequency of vibration is $f=\frac{1}{2\pi }\times \sqrt{\frac{2k}{m}}$ .

Explanation of Solution

Given info:

Mass of block $\text{= }m\text{ kg}$ .

Spring stiffness coefficient (for both spring) $\text{ = }k$ .

Formula used:

  $\begin{array}{l}\Sigma F_y=0\text{ (y is in equilibrium, net force is zero)}\\ \Sigma F_y=F_{\text{net}}\text{ (Newton 2nd law of motion)}\end{array}$

Here,

  $F_y$ is vertical force,

  $F_{\text{net}}$ is total net force.

For equilibrium the net force is equal to zero.

Calculation:

When object is completely coincided with $x$ axis and $y$ is at equilibrium, net force is zero in $y$ direction.

  $\begin{array}{l}\Sigma F_y=0\\ \Sigma F_y=F_A+F_B-mg=0\\ F_A+F_B=mg\text{ (Equation 1)}\end{array}$

When object is slightly below the x axis, then we apply the Newton's 2nd law of motion.

  $\begin{array}{l}\Sigma F_y=F_{net}=F_A{}^{\text{'}}+F_B{}^{\text{'}}-mg\\ F_A{}^{\text{'}}=F_A-kx\\ F_B{}^{\text{'}}=F_B-kx\\ F_{net}=(F_A+F_B)-2kx-mg\\ F_A+F_B=mg\text{ (from equation 1)}\\ F_{net}=-2kx\end{array}$

This looks like the general formula for the simple harmonic motion, here the effective spring constant is $2k$ .

  $\begin{array}{l}\text{Frequency of vibration }f=\frac{1}{2\pi }\times \sqrt{\frac{k_{\text{effective}}}{m}}\\ f=\frac{1}{2\pi }\times \sqrt{\frac{2k}{m}}\end{array}$

Conclusion:

Thus, the frequency of vibration is $f=\frac{1}{2\pi }\times \sqrt{\frac{2k}{m}}$ .