A certain simple pendulum has a period on the earth of 1.60 s. What is its period on the surface of Mars, where g = 3.71 m/s2?

Question: A certain simple pendulum has a period on the earth of 1.60 s. What is its period on the surface of Mars, where g = 3.71 m/s²?

Solution:

Given:

Time period on the earth surface of the pendulum (T_e)=1.60 Seconds

Acceleration due to gravity on Mars (g_m)=3.71m/s²

We Know

$T = 2\pi \sqrt {\frac{l}{g}} $.............(i)

Now,

Time period for earth (T_e)= $2\pi \sqrt {\frac{l}{{{g_e}}}} $.............(ii)

Time period for Mars (T_m)= $2\pi \sqrt {\frac{l}{{{g_m}}}} $.............(iii)

From Equation (ii) and (iii), We Get:

$\begin{array}{l}\frac{{{T_m}}}{{{T_e}}} = \sqrt {\frac{{{g_e}}}{{{g_m}}}} \\or,{\rm{ }}{T_m} = {T_e}\sqrt {\frac{{{g_e}}}{{{g_m}}}} \\\therefore {T_m} = 1.6 \times \sqrt {\frac{{9.8}}{{3.71}}}  = 2.6s\end{array}$

Thus, Time Period on the Mars Surface is 2.6Seconds