Question

Hanging from a horizontal beam are nine simple pendulums of the following lengths.

$$\begin{gathered} \left(\mathbf{a}\right)\mathbf{}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{,}\mathbf{} \\ \left(\mathbf{b}\right)\mathbf{}\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{,}\mathbf{} \\ \left(\mathbf{c}\right)\mathbf{}\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{,} \\ \mathbf{}\left(\mathbf{d}\right)\mathbf{}\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{,}\mathbf{} \\ \left(\mathbf{e}\right)\mathbf{}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{,} \\ \mathbf{}\left(\mathbf{f}\right)\mathbf{}\mathbf{2}\mathbf{.}\mathbf{8}\mathbf{,}\mathbf{} \\ \left(\mathbf{g}\right)\mathbf{}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{} \\ \left(\mathbf{h}\right)\mathbf{}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{,} \end{gathered}$$

$\mathbf{}\left(\mathbf{i}\right)\mathbf{}\mathbf{6}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{m}$Suppose the beam undergoes horizontal oscillations with angular frequencies in the range from$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{rad}\mathbf{/}\mathbf{s}$to$\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{rad}\mathbf{/}\mathbf{s}$. Which of the pendulums will be (strongly) set in motion?

Short answer

The pendulum D with length 0.80 m and pendulum E with length 1.2 m will set the resonance.

Step-by-step solution

Given

1. Length of pendulum A is,${\text{L}}_A=0.10\text{m}$
2. Length of pendulum B is,${\text{L}}_B=0.30\text{m}$
3. Length of pendulum C is,${\text{L}}_C=0.40m$
4. Length of pendulum D is,${\text{L}}_D=0.80\text{m}$
5. Length of pendulum E is,${\text{L}}_E=1.2\text{m}$
6. Length of pendulum F is,${\text{L}}_F=2.8\text{m}$
7. Length of pendulum G is,${\text{L}}_G=3.5\text{m}$
8. Length of pendulum H is,${\text{L}}_H=5.0\text{m}$
9. Length ofpendulum I is,${\text{L}}_I=6.2\text{m}$

Understanding the concept

Use the equation for angular frequency to calculate the angular frequency for each pendulum. Then according to the condition for resonance, the pendulums which have an angular frequency equal to that of the beam will set the resonance or will be strongly set in motion.

The angular frequency of the pendulum is given as-

$\mathbf{\omega }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{T}}$

The time period of oscillationcan be written as-

$\mathbf{T}\mathbf{=}\mathbf{2}\mathbf{\pi }\mathbf{}\sqrt{\frac{\mathbf{L}}{\mathbf{g}}}$

Here.Lis the length of the pendulum,is the acceleration due to gravity.

Write an expression for angular velocity

The equation for angular frequency is

$\omega =\frac{2\pi }{\text{T}}$

Here, the period of the oscillation is

$\text{T}=2\pi \sqrt{\frac{\text{L}}{\text{g}}}$

So, the equation for angular frequency becomes

$\omega =\sqrt{\frac{g}{L}}$

Calculate the angular velocities of the pendulums

$\begin{array}{l}{\omega }_A=\sqrt{\frac{g}{L_A}}\\ =\sqrt{\frac{9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}{0.10\text{\hspace{0.17em}m}}}\\ =9.9\text{\hspace{0.17em}rad}/\text{s}\end{array}$

role="math" localid="1661410601570" $\begin{array}{l}{\omega }_B=\sqrt{\frac{g}{L_B}}\\ =\sqrt{\frac{9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}{0.30\text{\hspace{0.17em}m}}}\\ =5.72\text{\hspace{0.17em}rad}/\text{s}\end{array}$

role="math" localid="1661410585924" $\begin{array}{l}{\omega }_C=\sqrt{\frac{g}{L_C}}\\ =\sqrt{\frac{9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}{0.40\text{\hspace{0.17em}m}}}\\ =4.95\text{\hspace{0.17em}rad}/\text{s}\end{array}$

$\begin{array}{l}{\omega }_D=\sqrt{\frac{g}{L_D}}\\ =\sqrt{\frac{9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}{0.80\text{\hspace{0.17em}m}}}\\ =3.5\text{\hspace{0.17em}rad}/\text{s}\end{array}$

$\begin{array}{l}{\omega }_E=\sqrt{\frac{g}{L_E}}\\ =\sqrt{\frac{9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}{1.2\text{\hspace{0.17em}m}}}\\ =2.86\text{\hspace{0.17em}rad}/\text{s}\end{array}$

$\begin{array}{l}{\omega }_F=\sqrt{\frac{g}{L_F}}\\ =\sqrt{\frac{9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}{2.8\text{\hspace{0.17em}m}}}\\ =1.87\text{\hspace{0.17em}rad}/\text{s}\end{array}$

$\begin{array}{l}{\omega }_G=\sqrt{\frac{g}{L_G}}\\ =\sqrt{\frac{9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}{3.5\text{\hspace{0.17em}m}}}\\ =1.67\text{\hspace{0.17em}rad}/\text{s}\end{array}$

$\begin{array}{l}{\omega }_H=\sqrt{\frac{g}{L_H}}\\ =\sqrt{\frac{9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}{5.0\text{\hspace{0.17em}m}}}\\ =1.14\text{\hspace{0.17em}rad}/\text{s}\end{array}$

$\begin{array}{l}{\omega }_I=\sqrt{\frac{g}{L_I}}\\ =\sqrt{\frac{9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}{6.2\text{\hspace{0.17em}m}}}\\ =1.26\text{\hspace{0.17em}rad}/\text{s}\end{array}$

From these all values, the only ${\omega }_D=3.5\text{rad}/\text{s}$and${\omega }_E=2.86\text{rad}/\text{s}$are in the range of the frequency of the beam that is$2.00\text{\hspace{0.17em}rad}/\text{s}$and$4.00\text{\hspace{0.17em}rad}/\text{s}$

So, we can say that pendulum D with length 0.80 m and pendulum E with length 1.2 m will be strongly set in motion.