Chapter 14, Problem 92P

(a)

To determine

The time period of oscillating system.

Explanation of Solution

Introduction:

Time period of oscillating system is the time required by system to complete one oscillation.

The free body diagram of system is drawn below.

  

Here, $Mg$ is the weight of bob, $kx$ is the spring force, $L$ is the length of pendulum, $T$ is the tension in string and $\theta$ is the angle of string with the vertical.

Write the expression for time period of the pendulum.

  $T_p=\frac{2\pi }{\omega }$ …… (1)

Here, $T_p$ is time period and $\omega$ is angular frequency of pendulum.

Write the expression for resultant force in horizontal direction.

  ${\displaystyle \sum F_x=0}$

Substitute $-kx-T\sin \theta -M{a}_x$ for ${\displaystyle \sum F_x}$ in above expression.

  $-kx-T\sin \theta -M{a}_x=0$ …… (2)

Here, $k$ is the spring constant, $x$ is the extension in length, $T$ is the tension, $M$ is the mass of bob, $a_x$ is the acceleration and $\theta$ is the angle with the vertical.

Write the expression for resultant force in horizontal direction.

  ${\displaystyle \sum F_y=0}$

Substitute $T\cos \theta -Mg$ for ${\displaystyle \sum F_y}$ in above expression.

  $T\cos \theta -Mg=0$

Here, $g$ is the gravitational acceleration.

Rearrange the above expression in term of $T$ .

  $T=\frac{Mg}{\cos \theta }$

Substitute $\frac{Mg}{\cos \theta }$ for $T$ in equation (1) and simplify.

  $-kx-Mg\tan \theta -M{a}_x=0$

Substitute $L\theta$ for $x$ and $L\frac{d^2\theta }{d{t}^2}$ for $a_x$ in above expression.

  $-kL\theta -Mg\tan \theta -ML\frac{d^2\theta }{d{t}^2}=0$

Rearrange the above expression in term of $\frac{d^2\theta }{d{t}^2}$ for $\theta \ll 1$ .

  $\frac{d^2\theta }{d{t}^2}=-\left(\frac{k}{M}+\frac{g}{L}\right)\theta$

Compare the above expression with the equation of differential equation of S.H.M. $d^2\theta /d{t}^2=-{\omega }^2\theta$ .

The value of angular frequency is $\omega =\sqrt{\frac{k}{M}+\frac{g}{L}}$

Substitute $\sqrt{\frac{k}{M}+\frac{g}{L}}$ for $\omega$ in equation (1).

  $T_p=\frac{2\pi }{\sqrt{\frac{k}{M}+\frac{g}{L}}}$ …… (3)

Conclusion:

Thus, the time period of oscillating system is $T_p=\frac{2\pi }{\sqrt{\frac{k}{M}+\frac{g}{L}}}$ .

(b)

To determine

The value of force constant.

Explanation of Solution

Given:

The mass of bob is $1.00\text{kg}$ .

The first time period of the pendulum is $2.00\text{s}$ .

The second time period of the pendulum is $1.00\text{s}$ .

Formula:

The free body diagram of system is drawn below.

  

Here, $Mg$ is the weight of bob, $kx$ is the spring force, $L$ is the length of pendulum, $T$ is the tension in string and $\theta$ is the angle of string with the vertical.

Write the expression for time period of the pendulum.

  $T_p=\frac{2\pi }{\omega }$ …… (1)

Here, $T_p$ is time period and $\omega$ is angular frequency of pendulum.

Write the expression for resultant force in horizontal direction.

  ${\displaystyle \sum F_x=0}$

Substitute $-kx-T\sin \theta -M{a}_x$ for ${\displaystyle \sum F_x}$ in above expression.

  $-kx-T\sin \theta -M{a}_x=0$ …… (2)

Here, $k$ is the spring constant, $x$ is the extension in length, $T$ is the tension, $M$ is the mass of bob, $a_x$ is the acceleration and $\theta$ is the angle with the vertical.

Write the expression for resultant force in horizontal direction.

  ${\displaystyle \sum F_y=0}$

Substitute $T\cos \theta -Mg$ for ${\displaystyle \sum F_y}$ in above expression.

  $T\cos \theta -Mg=0$

Here, $g$ is the gravitational acceleration.

Rearrange the above expression in term of $T$ .

  $T=\frac{Mg}{\cos \theta }$

Substitute $\frac{Mg}{\cos \theta }$ for $T$ in equation (1) and simplify.

  $-kx-Mg\tan \theta -M{a}_x=0$

Substitute $L\theta$ for $x$ and $L\frac{d^2\theta }{d{t}^2}$ for $a_x$ in above expression.

  $-kL\theta -Mg\tan \theta -ML\frac{d^2\theta }{d{t}^2}=0$

Rearrange the above expression in term of $\frac{d^2\theta }{d{t}^2}$ for $\theta \ll 1$ .

  $\frac{d^2\theta }{d{t}^2}=-\left(\frac{k}{M}+\frac{g}{L}\right)\theta$

Compare the above expression with the equation of differential equation of S.H.M. $d^2\theta /d{t}^2=-{\omega }^2\theta$ .

The value of angular frequency is $\omega =\sqrt{\frac{k}{M}+\frac{g}{L}}$

Substitute $\sqrt{\frac{k}{M}+\frac{g}{L}}$ for $\omega$ in equation (1).

  $T_p=\frac{2\pi }{\sqrt{\frac{k}{M}+\frac{g}{L}}}$ …… (3)

Calculation:

For $k=0$ , $T_p=2.00\text{s}$ :

Substitute $2.00\text{s}$ for $T_p$ , $0$ for $k$ , $1.00\text{kg}$ for $M$ in equation (3).

  $\begin{array}{c}2.00\text{s}=\frac{2\pi }{\sqrt{\frac{0}{1.00\text{kg}}+\frac{g}{L}}}\\ \frac{g}{L}={\pi }^2\end{array}$

For $T_p=1.00\text{s}$ :

Substitute $1.00\text{s}$ for $T_p$ , ${\pi }^2$ for $\frac{g}{L}$ and $1.00\text{kg}$ for $M$ in equation (3).

  $\begin{array}{c}1.00\text{s}=\frac{2\pi }{\sqrt{\frac{k}{1.00\text{kg}}+{\pi }^2}}\\ k=29.6\text{N}/\text{m}\end{array}$

Conclusion:

Thus, the value of force constant is $29.6\text{N}/\text{m}$ .