Chapter 14, Problem 50A

To determine

To Rank: The pendulums according to period, form least to greatest.

Answer to Problem 50A

  $\text{A}\text{=}\text{B}<\text{C}\text{=}\text{D}$

Explanation of Solution

Formula:

Time period of simple pendulum is

  $T=2\pi \sqrt{\frac{L}{g}}$  ...... (1)

Here, $L$ is the length of the string and $g$ is the acceleration due to gravity.

Calculation:

For option A, the length of the pendulum is $10\text{cm}$ and mass is $0.25\text{kg}$

Substitute $10\text{cm}$ for $L$ and $9.\text{8}{\text{m/s}}^{\text{2}}$ for $g$ in equation (1)

  $\begin{array}{l}T=2\pi \sqrt{\frac{L}{g}}\\ T=6.28\sqrt{\frac{0.1}{9.8}}\\ T=6.28\times 0.1\\ T_1=0.628\mathrm{s}\end{array}$

For option B here the length of the pendulum is $10\text{cm}$ and mass is $0.35\text{kg}$ .

Substitute $10\text{cm}$ for $L$ and $9.\text{8}{\text{m/s}}^{\text{2}}$ for $g$ in equation (1)

  $\begin{array}{l}T=2\pi \sqrt{\frac{L}{g}}\\ T=6.28\sqrt{\frac{0.1}{9.8}}\\ T=6.28\times 0.1\\ T_1=0.628\mathrm{s}\end{array}$

For option C here the length of the pendulum is $20\text{cm}$ and mass is $0.25\text{kg}$

Substitute $20\text{cm}$ for $L$ and $9.\text{8}{\text{m/s}}^{\text{2}}$ for $g$ in equation (1)

  $\begin{array}{l}T=2\pi \sqrt{\frac{L}{g}}\\ T=6.28\sqrt{\frac{0.2}{9.8}}\\ T=6.28\times 0.141\\ T=0.88\mathrm{s}\end{array}$

For option D here the length of the pendulum is $20\text{cm}$ and mass is $0.35\text{kg}$

Substitute $20\text{cm}$ for $L$ and $9.\text{8}{\text{m/s}}^{\text{2}}$ for $g$ in equation (1)

  $\begin{array}{l}T=2\pi \sqrt{\frac{L}{g}}\\ T=6.28\sqrt{\frac{0.2}{9.8}}\\ T=6.28\times 0.141\\ T=0.88\mathrm{s}\end{array}$

Period of pendulum doesn’t depend on mass of the pendulum. It depends on length of the pendulum. Therefore, its arrangement is $\text{A}\text{=}\text{B}<\text{C}\text{=}\text{D}$ .

Conclusion:

Hence, the arrangement is $\text{A}\text{=}\text{B}<\text{C}\text{=}\text{D}$ .