91Ó°ÊÓ

If the amplitude of oscillations is small and the effect of friction can be ignored, then a simple pendulum is said to achieve the simple harmonic motion.

The time required to perform one cycle of oscillation of the pendulum is termed as the time period.

The time period of simple pendulum on the Earth is, \({T_{{\rm{Earth}}}} = 1.85\;{\rm{s}}\).

The acceleration due to gravity on the Earth is, \({g_{{\rm{Earth}}}} = 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\).

The acceleration due to gravity on the Mars is, \({g_{{\rm{Mars}}}} = 0.37{g_{{\rm{Earth}}}}\).

The time period of oscillation of simple pendulum for small amplitudes is:

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

Here, g is acceleration due to gravity and l is the length of the simple pendulum.

Thus, the time period of oscillation of simple pendulum on the Earth can be written as:

\({T_{{\rm{Earth }}}} = 2\pi \sqrt {\frac{l}{{{g_{{\rm{Earth }}}}}}} \)

Since length of the simple pendulum does not change on Mars, thus the time period of simple pendulum on the Mars is,

\({T_{{\rm{Mars }}}} = 2\pi \sqrt {\frac{l}{{{g_{{\rm{Mars }}}}}}} \)

The ratio of time period on the Mars to the time period on the Earth is,

\(\begin{aligned}{c}\frac{{{T_{{\rm{Mars }}}}}}{{{T_{{\rm{Earth }}}}}} &= \frac{{2\pi \sqrt {\frac{l}{{{g_{{\rm{Mars}}}}}}} }}{{2\pi \sqrt {\frac{l}{{{g_{{\rm{Earth }}}}}}} }}\\ &= \sqrt {\frac{{{g_{{\rm{Earth }}}}}}{{{g_{{\rm{Mars }}}}}}} \end{aligned}\)

Thus, the time period on the Mars is:

\({T_{{\rm{Mars }}}} = {T_{{\rm{Earth }}}}\sqrt {\frac{{{g_{{\rm{Earth }}}}}}{{{g_{{\rm{Mars }}}}}}} \)

On substituting the given values in the above expression.

\(\begin{aligned}{c}{T_{{\rm{Mars }}}} &= {T_{{\rm{Earth }}}}\sqrt {\frac{{{g_{{\rm{Earth }}}}}}{{0.37{g_{{\rm{Earth }}}}}}} \\ &= (1.85\;{\rm{s}})\sqrt {\frac{1}{{0.37}}} \\ &= 3.0399\;{\rm{s}}\\ \approx 3.0\;{\rm{s}}\end{aligned}\)

Hence, the period on the Mars is 3.0 s.