Chapter 11, Problem 30P

To determine

To Find: The period of pendulum on Mars.

Answer to Problem 30P

The period of pendulum on Mars is $1.32\text{ s}$ .

Explanation of Solution

Given info:

Time period of earth $T_{\text{Earth}}=0.8\text{ s}$

Gravity on Mars $g_{\text{mars}}=0.37\times g$

Length of pendulum on Earth and Mars assumed same.

Formula used:

Write the expression for time period.

  $T=2\pi \times \sqrt{\frac{l}{g}}$

Here,

  $T$ istime period of pendulum,

  $l$ is the length of pendulum,

  $g$ is the acceleration due to gravity on earth.

Calculation:

The expression for time period of Mars.

  $T_{\text{Mars}}=2\pi \times \sqrt{\frac{l}{g_{\text{Mars}}}}$

The expression for time period of Earth.

  $T_{\text{Earth}}=2\pi \times \sqrt{\frac{l}{g_{\text{Earth}}}}$

Take the ratio of period on Mars to Earth, the period of Mars is calculated by following way:

  $\begin{array}{l}\frac{T_{\text{Mars}}}{T_{\text{Earth}}}=\frac{2\pi \times \sqrt{\frac{l}{g_{\text{Mars}}}}}{2\pi \times \sqrt{\frac{l}{g_{\text{Earth}}}}}=\sqrt{\frac{g_{\text{Earth}}}{g_{\text{Mars}}}}\\ \frac{T_{\text{Mars}}}{T_{\text{Earth}}}=\sqrt{\frac{9.8}{0.37\times 9.8}}=1.64\\ T_{\text{Mars}}=1.64\times T_{\text{Earth}}=1.64\times 0.8=1.32\text{ s}\end{array}$

Conclusion:

Thus, period of pendulum on Mars is $1.32\text{ s}$ .