Ncert Exercises & Solutions

The length of a second's pendulum on the surface of earth is 1 m. What will be the length of a second's pendulum on the moon?

Hint: The time period remains the same.

Step 1: Find the time period on the surface of the earth.

A second's pendulum means a simple pendulum having a time period, T=2 s

For a simple pendulum,

T = 2 π \sqrt{\frac{l}{g}}

$T = 2 π \sqrt{\frac{l}{g}}$

where l =length of the pendulum and g= acceleration due to gravity on surface of the earth.

T_{e} = 2 π \sqrt{\frac{l_{e}}{g_{e}}} . . . (i)

$T_{e} = 2 π \sqrt{\frac{l_{e}}{g_{e}}} . . . (i)$

Step 2: Find the time period on the surface of the moon.

On the surface of the moon,

T_{m} = 2 π \sqrt{\frac{l_{m}}{g_{m}}}

$T_{m} = 2 π \sqrt{\frac{l_{m}}{g_{m}}}$

Step 3: Find the length of the pendulum on the surface of the moon.

∴ \frac{T_{e}}{T_{m}} = \frac{2 π}{2 π} \sqrt{\frac{l_{e}}{g_{e}}} \times \sqrt{\frac{g_{m}}{l_{m}}}

$∴ \frac{T_{e}}{T_{m}} = \frac{2 π}{2 π} \sqrt{\frac{l_{e}}{g_{e}}} \times \sqrt{\frac{g_{m}}{l_{m}}}$

T_{e} = T_{m}

$T_{e} = T_{m}$ to maintain the second's pendulum time period.

∴ 1 = \sqrt{\frac{l_{e}}{l_{m}} \times \frac{g_{m}}{g_{e}}}

$∴ 1 = \sqrt{\frac{l_{e}}{l_{m}} \times \frac{g_{m}}{g_{e}}}$ ...(ii)

But the acceleration due to gravity at moon is 1/6 of the acceleration due to gravity at earth, i.e.,

g_{m} = \frac{g_{e}}{6}

$g_{m} = \frac{g_{e}}{6}$

Squaring Eq.(iii) and putting this value,

1 = \frac{l_{e}}{l_{m}} \times \frac{g_{e} / 6}{g_{e}} = \frac{l_{e}}{l_{m}} \times \frac{1}{6}

$1 = \frac{l_{e}}{l_{m}} \times \frac{g_{e} / 6}{g_{e}} = \frac{l_{e}}{l_{m}} \times \frac{1}{6}$

\Rightarrow \frac{l_{e}}{6 l_{m}} = 1 \Rightarrow 6 l_{m} = l_{e}

$\Rightarrow \frac{l_{e}}{6 l_{m}} = 1 \Rightarrow 6 l_{m} = l_{e}$

\Rightarrow l_{m} = \frac{1}{6} l_{e} = \frac{1}{6} \times 1 = \frac{1}{6} m

$\Rightarrow l_{m} = \frac{1}{6} l_{e} = \frac{1}{6} \times 1 = \frac{1}{6} m$