Chapter 13, Problem 40P

A simple pendulum is 5.00 in long. (a) What is the period of simple harmonic motion for this pendulum if it is located in an elevator accelerating upward at 5.00 m/s²? (b) What is its period if the elevator is accelerating downward at 5.00 m/s²? (c) What is the period of simple harmonic motion for the pendulum if it is placed in a truck that is accelerating horizontally at 5.00 m/s²?

To determine

The period of the simple harmonic motion of the pendulum which is in an elevator.

Answer to Problem 40P

The time period of the pendulum is $3.65\text{s}$.

Explanation of Solution

Given info: The length of the pendulum is $5.00\text{m}$. The elevator is accelerating upward at $5.00\text{m}\cdot {\text{s}}^{\text{-2}}$.

The free body diagram of the pendulum,

From the figure,

$T-mg=ma$ (1)

• $T$ is the tension on the string
• $m$ is the mass of the object
• $g$ is the free fall acceleration
• $a$ is the acceleration of the elevator

From (1)

$\begin{array}{c}T=m\left(g+a\right)\\ =m{g}_{apparent}\end{array}$

• $g_{apparent}$ is the apparent free fall acceleration

Choose the downward direction to be the positive direction. Apparent free fall acceleration of the pendulum in the elevator is,

$\begin{array}{c}{g}_{apparent}=g+a\\ =9.80\text{m}\cdot {\text{s}}^{\text{-2}}+5.00\text{m}\cdot {\text{s}}^{\text{-2}}\\ =14.80\text{m}\cdot {\text{s}}^{\text{-2}}\end{array}$

The period of a pendulum is defined by,

$T=2\pi \sqrt{\frac{L}{g_{apparent}}}$

• $T$ is the period
• $L$ is the length of the string
• $g_{apparent}$ is the apparent free fall acceleration

Substituting $L=5.00\text{m}$ and $g_{apparent}=14.80\text{m}\cdot {\text{s}}^{\text{-2}}$,

$\begin{array}{c}T=2\pi \sqrt{\frac{\left(5.00\text{m}\right)}{\left(14.80\text{m}\cdot {\text{s}}^{\text{-2}}\right)}}\\ =3.65\text{s}\end{array}$

Conclusion: The time period of the pendulum is $3.65\text{s}$.

To determine

The period of the simple harmonic motion of the pendulum this is in an elevator.

Answer to Problem 40P

The time period of the pendulum is $6.41\text{s}$.

Explanation of Solution

Given info: The length of the pendulum is $5.00\text{m}$. The elevator is accelerating downward at $5.00\text{m}\cdot {\text{s}}^{\text{-2}}$.

The free body diagram of the pendulum,

From the figure,

$mg-T=ma$ (1)

• $T$ is the tension on the string
• $m$ is the mass of the object
• $g$ is the free fall acceleration
• $a$ is the acceleration of the elevator

From (1)

$\begin{array}{c}T=m\left(g-a\right)\\ =m{g}_{apparent}\end{array}$

• $g_{apparent}$ is the apparent free fall acceleration

Choose the downward direction to be the positive direction. Apparent free fall acceleration of the pendulum in the elevator is,

$\begin{array}{c}{g}_{apparent}=g-a\\ =9.80\text{m}\cdot {\text{s}}^{\text{-2}}-5.00\text{m}\cdot {\text{s}}^{\text{-2}}\\ =4.80\text{m}\cdot {\text{s}}^{\text{-2}}\end{array}$

The period of a pendulum is defined by,

$T=2\pi \sqrt{\frac{L}{g_{apparent}}}$

• $T$ is the period
• $L$ is the length of the string
• $g_{apparent}$ is the apparent free fall acceleration

Substituting $L=5.00\text{m}$ and $g_{apparent}=4.80\text{m}\cdot {\text{s}}^{\text{-2}}$,

$\begin{array}{c}T=2\pi \sqrt{\frac{\left(5.00\text{m}\right)}{\left(4.80\text{m}\cdot {\text{s}}^{\text{-2}}\right)}}\\ =6.41\text{s}\end{array}$

Conclusion: The time period of the pendulum is $6.41\text{s}$.

To determine

The period of a simple pendulum this is in an accelerating truck.

Answer to Problem 40P

The period of the pendulum is $4.24\text{s}$.

Explanation of Solution

Given info: The length of the pendulum is $5.00\text{m}$. The truck is accelerating horizontally at $5.00\text{m}\cdot {\text{s}}^{\text{-2}}$.

The figure showing the apparent free fall acceleration,

From the figure, using triangular law of vector addition, the magnitude of $g_{apparent}$ will be,

$\begin{array}{c}{g}_{apparent}=\sqrt{{\left(5.00\text{m}\cdot {\text{s}}^{\text{-2}}\right)}^2+\left(9.80\text{m}\cdot {\text{s}}^{\text{-2}}\right)}\\ =11.0\text{m}\cdot {\text{s}}^{\text{-2}}\end{array}$

The period of a pendulum is defined by,

$T=2\pi \sqrt{\frac{L}{g_{apparent}}}$

• $T$ is the period
• $L$ is the length of the string
• $g_{apparent}$ is the apparent free fall acceleration

Substituting $L=5.00\text{m}$ and $g_{apparent}=11.0\text{m}\cdot {\text{s}}^{\text{-2}}$,

$\begin{array}{c}T=2\pi \sqrt{\frac{\left(5.00\text{m}\right)}{\left(11.0\text{m}\cdot {\text{s}}^{\text{-2}}\right)}}\\ =4.24\text{s}\end{array}$

Conclusion: The period of the pendulum on the accelerating truck is $4.24\text{s}$.