Chapter 19.5, Problem 19.137P

To determine

The minimum tension occur in each spring.

Answer to Problem 19.137P

The minimum tension occur in each spring is $8.82\text{N}$.

Explanation of Solution

Given information:

The mass of the block $A$ is $2.4\text{kg}$, the mass of the block $B$ is $0.9\text{kg}$, the spring constant of each spring is $180\text{N}/\text{m}$, the damping coefficient is $7.5\text{N}\cdot \text{s}/\text{m}$

The figure illustrates the free positions and extreme position.

Figure-(1)

Write the balanced equation of spring force and weight.

$\begin{array}{l}2k{\Delta }_1=\left(m_{\text{A}}+m_{\text{B}}\right)g\\ {\Delta }_1=\frac{\left(m_{\text{A}}+m_{\text{B}}\right)g}{2k}\end{array}$ ..... (I)

Here, the mass of the block $A$ is $m_{\text{A}}$, the mass of the block $B$ is $m_{\text{B}}$, the spring constant of each spring is $k$, the elongation of spring at equilibrium position is ${\Delta }_1$.

Write the balanced equation of spring force and weight after cutting the cord.

$\begin{array}{l}2k{\Delta }_2=\left(m_{\text{B}}\right)g\\ {\Delta }_2=\frac{\left(m_{\text{A}}\right)g}{2k}\end{array}$ ...... (II)

Here, the elongation of spring after cutting the cord is ${\Delta }_2$.

Write the expression of displacement of block $A$ from its mean position.

$x=x_0{e}^{-\zeta {\omega }_{\text{n}}t}\sin \left({\omega }_{\text{d}}t+\varphi \right)$ ...... (III)

Here, the maximum amplitude is $x_0$, the damping factor is $\zeta$, the angular damping frequency is ${\omega }_{\text{d}}$, the time period of oscillation is $t$, the phase difference is $\varphi$.

Write the expression of initial displacement of block $A$ at $t=0\text{s}$.

$x_1={\Delta }_1-{\Delta }_2$ ...... (IV)

Here, the displacement of block $A$ at $t=0\text{s}$ is $x_1$.

Differentiate the Equation (III) with respect to time.

$\begin{array}{l}x=x_0{e}^{-\zeta {\omega }_{\text{n}}t}\sin \left({\omega }_{\text{d}}t+\varphi \right)\\ \frac{dx}{dt}=\frac{d\left(x_0{e}^{-\zeta {\omega }_{\text{n}}t}\sin \left({\omega }_{\text{d}}t+\varphi \right)\right)}{dt}\\ V=x_0\left[{\omega }_{\text{d}}{e}^{-\zeta {\omega }_{\text{n}}t}\cos \left({\omega }_{\text{d}}t+\varphi \right)+\sin \left({\omega }_{\text{d}}t+\varphi \right)\left(-\zeta {\omega }_{\text{n}}\right)\right]\end{array}$ ...... (V)

Substitute $0\text{s}$ for $t$ as initial condition in Equation (V).

$V_{t=0}=x_0\left[{\omega }_{\text{d}}\cos \left(\varphi \right)+\sin \left(\varphi \right)\left(-\zeta {\omega }_{\text{n}}\right)\right]$ ...... (VI)

Write the expression of natural frequency of vibration of block $A$.

${\omega }_{\text{n}}=\sqrt{\frac{k_{\text{eq}}}{m_{\text{A}}}}$ ...... (VII)

Here, the natural frequency of vibrations is ${\omega }_{\text{n}}$, the equivalent spring constant is $k_{\text{eq}}$.

Write the expression of damping factor.

$\zeta =\frac{c}{2m_{\text{A}}{\omega }_{\text{n}}}$ ...... (VIII)

Here, the damping factor is $\zeta$, the damping coefficient is $c$.

Write the expression of damped frequency in terms of natural frequency.

${\omega }_{\text{d}}={\omega }_{\text{n}}\sqrt{1-{\zeta }^2}$ ....... (IX)

The velocity of the block $A$ becomes $0$ after the start when ${\omega }_{\text{d}}t=\pi$.

Write the expression of time period of oscillations.

$t=\frac{\pi }{{\omega }_{\text{d}}}$ ...... (X)

Here, the time period of oscillations is $t$.

Write the expression for minimum stretch.

$\begin{array}{c}S={\Delta }_2+x\\ =\frac{\left(m_{\text{A}}\right)g}{2k}+x\end{array}$ ...... (XI)

Here, the minimum stretch in the spring is $S$.

Write the expression of minimum tension in the spring.

$T=K\times S$ ...... (XII)

Here, the minimum tension in the spring is $T$.

Calculation:

Substitute $\frac{\left(m_{\text{A}}+m_{\text{B}}\right)g}{2k}$ for ${\Delta }_1$ and $\frac{\left(m_{\text{A}}\right)g}{2k}$ for ${\Delta }_2$ in Equation (IV).

$\begin{array}{c}{x}_1=\frac{\left(m_{\text{A}}+m_{\text{B}}\right)g}{2k}-\frac{\left(m_{\text{A}}\right)g}{2k}\\ =\frac{\left(m_{\text{A}}\right)g}{2k}+\frac{m_{\text{B}}g}{2k}-\frac{\left(m_{\text{A}}\right)g}{2k}\\ =\frac{m_{\text{B}}g}{2k}\end{array}$ ...... (XIII)

Substitute $0\text{s}$ for $t$ in Equation (III).

$x_1=x_0\sin \left(\varphi \right)$ ...... (XIV)

Substitute $0\text{m}/\text{s}$ for $V$ and ${\omega }_{\text{n}}\sqrt{1-{\zeta }^2}$ for ${\omega }_{\text{d}}$ in Equation (VI).

$\begin{array}{l}0=x_0\left[{\omega }_{\text{d}}\cos \left(\varphi \right)+\sin \left(\varphi \right)\left(-\zeta {\omega }_{\text{n}}\right)\right]\\ \sin \left(\varphi \right)\left(\zeta {\omega }_{\text{n}}\right)={\omega }_{\text{d}}\cos \left(\varphi \right)\\ \sin \left(\varphi \right)\left(\zeta {\omega }_{\text{n}}\right)={\omega }_{\text{n}}\cos \left(\varphi \right)\sqrt{1-{\zeta }^2}\\ \tan \varphi =\frac{\sqrt{1-{\zeta }^2}}{\zeta }\end{array}$ ...... (XV)

Substitute $360\text{N}/\text{m}$ for $k_{\text{eq}}$, $2.4\text{kg}$ for $m_{\text{A}}$ in Equation (VII).

$\begin{array}{c}{\omega }_{\text{n}}=\sqrt{\frac{360\text{N}/\text{m}}{2.4\text{kg}}}\\ =\sqrt{150{\text{rad}}^2/{\text{s}}^2}\\ =12.247\text{rad}/\text{s}\end{array}$

Substitute $12.247\text{rad}/\text{s}$ for ${\omega }_{\text{n}}$, $2.4\text{kg}$ for $m_{\text{A}}$, $7.5\text{N}\cdot \text{s}/\text{m}$ for $c$ in Equation (VIII).

$\begin{array}{c}\zeta =\frac{7.5\text{N}\cdot \text{s}/\text{m}}{2\left(2.4\text{kg}\right)\left(12.247\text{rad}/\text{s}\right)}\\ =\frac{7.5\text{N}\cdot \text{s}/\text{m}}{58.785\text{kg}\cdot \text{rad}/\text{s}}\\ =0.127\end{array}$

Substitute $0.127$ for $\zeta$ in Equation (XV).

$\begin{array}{l}\tan \varphi =\frac{\sqrt{1-{\left(0.127\right)}^2}}{0.127}\\ \tan \varphi =7.810\\ \varphi ={\tan }^{-1}\left(7.810\right)\\ \varphi =82.703°\end{array}$

Equate the Equations (XIII) and (XIV).

$\begin{array}{l}{x}_0\sin \left(\varphi \right)=\frac{m_{\text{B}}g}{2k}\\ \end{array}$ ...... (XVI)

Substitute $82.703°$ for $\varphi$, $0.9\text{kg}$ for $m_{\text{B}}$, $180\text{N}/\text{m}$ for $k$, $9.81\text{m}/{\text{s}}^2$ for $g$ in Equation (XVI).

$\begin{array}{l}{x}_0\sin \left(82.703°\right)=\frac{0.9\text{kg}\left(9.81\text{m}/{\text{s}}^2\right)}{2\left(180\text{N}/\text{m}\right)}\\ 0.9919x_0=0.0245\text{m}\\ x_0=0.02472\text{m}\end{array}$

Substitute ${\omega }_{\text{n}}\sqrt{1-{\zeta }^2}$ for ${\omega }_{\text{d}}$, $82.703°$ for $\varphi$, $0.127$ for $\zeta$ in Equation (V).

$\begin{array}{l}V=x_0\left[{\omega }_{\text{n}}\sqrt{1-{\zeta }^2}{e}^{-\zeta {\omega }_{\text{n}}t}\cos \left({\omega }_{\text{d}}t+82.703°\right)+\sin \left({\omega }_{\text{d}}t+82.703°\right)\left(-\zeta {\omega }_{\text{n}}\right)\right]\\ V=x_0{\omega }_{\text{n}}{e}^{-\zeta {\omega }_{\text{n}}t}\left[\sqrt{1-{\zeta }^2}\cos \left({\omega }_{\text{d}}t+82.703°\right)+\sin \left({\omega }_{\text{d}}t+82.703°\right)\left(-\zeta \right)\right]\\ V=x_0{\omega }_{\text{n}}{e}^{-\zeta {\omega }_{\text{n}}t}\left[\cos \left({\omega }_{\text{d}}t+82.703°+{\tan }^{-1}\left(\frac{0.127}{\sqrt{1-{\left(0.127\right)}^2}}\right)\right)\right]\\ V=x_0{\omega }_{\text{n}}{e}^{-\zeta {\omega }_{\text{n}}t}\left[\cos \left({\omega }_{\text{d}}t+82.703°+7.33°\right)\right]\end{array}$

$\begin{array}{l}V=x_0{\omega }_{\text{n}}{e}^{-\zeta {\omega }_{\text{n}}t}\left[\cos \left({\omega }_{\text{d}}t+90°\right)\right]\\ V=-x_0{\omega }_{\text{n}}{e}^{-\zeta {\omega }_{\text{n}}t}\sin \left({\omega }_{\text{d}}t\right)\end{array}$

Substitute ${\omega }_{\text{n}}\sqrt{1-{\zeta }^2}$ for ${\omega }_{\text{d}}$, $0.127$ for $\zeta$, $12.247\text{rad}/\text{s}$ for ${\omega }_{\text{n}}$ in Equation (X).

$\begin{array}{c}t=\frac{\pi }{12.247\text{rad}/\text{s}\sqrt{1-{\left(0.127\right)}^2}}\\ =\frac{\pi }{12.147}\\ =0.258\text{s}\end{array}$

Substitute ${\omega }_{\text{n}}\sqrt{1-{\zeta }^2}$ for ${\omega }_{\text{d}}$, $0.127$ for $\zeta$, $12.247\text{rad}/\text{s}$ for ${\omega }_{\text{n}}$, $0.02472\text{m}$ for $x_0$, $\pi$ for ${\omega }_{\text{d}}t$, $82.703°$ for $\varphi$ in Equation (III).

$\begin{array}{c}x=\left[\left(0.02472\text{m}\right)e^{-\left(0.127\right)\left(12.247\text{rad}/\text{s}\right)\left(0.258\text{s}\right)}\sin \left(\pi +82.703°\right)\right]\\ =0.01654\sin 262.703°\\ =-0.0164\text{m}\end{array}$

Substitute $-0.0164\text{m}$ for $x$, $2.4\text{kg}$ for $m_{\text{A}}$, $180\text{N}/\text{m}$ for $k$, $9.81\text{m}/{\text{s}}^2$ for $g$ in Equation (XI).

$\begin{array}{c}S=\frac{\left(2.4\text{kg}\right)9.81\text{m}/{\text{s}}^2}{2\left(180\text{N}/\text{m}\right)}+\left(-0.0164\text{m}\right)\\ =0.0654\text{m}-0.0164\text{m}\\ =0.049\text{m}\end{array}$

Substitute $180\text{N}/\text{m}$ for $k$, $0.049\text{m}$ for $S$ in Equation (XII).

$\begin{array}{l}T=\left(180\text{N}/\text{m}\right)\times \left(0.049\text{m}\right)\\ T=8.82\text{N}\end{array}$

Conclusion:

The minimum tension occur in each spring is $8.82\text{N}$.