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The length of a second's pendulum on the surface of earth is $1 \mathrm{~m}$. What will be the length of a second's pendulum on the moon?
Solution:
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Verified Answer
A simple pendulum having time period $T=2 \mathrm{~s}$ is called second pendulum.
For a simple pendulum , $T=2 \pi \sqrt{\frac{l}{g}}$
where, $l=$ length of the pendulum and $g=$ acceleration due to gravity on surface of the earth.
$$
\begin{aligned}
&T_e=2 \pi \sqrt{\frac{l_e}{g_e}} \\
&T_e^2=4 \pi^2\left(\frac{l_e}{g_e}\right)
\end{aligned}
$$
On the surface of the moon, the time period
$$
\begin{aligned}
&\left(T_m\right)=2 \pi \sqrt{\frac{l_m}{g_m}} \Rightarrow T_m^2=4 \pi^2\left(\frac{l_m}{g_m}\right) \\
&\left(\frac{T_e}{T_m}\right)^2=\frac{4 \pi^2}{4 \pi^2}\left(\frac{l_e}{g_e}\right) \times\left(\frac{g_m}{l_m}\right)
\end{aligned}
$$
Dividing eq. (i) from (ii),
$$
\left(\frac{T_e}{T_m}\right)=\sqrt{\frac{l_e}{l_m} \times \frac{g_m}{g_e}}
$$
$T_e=T_m=2$ sec to maintain the second's pendulum time period.
$$
1=\sqrt{\frac{l_e}{l_m} \times \frac{g_m}{g_e}}
$$
But the acceleration due to gravity at moon is $1 / 6$ of the acceleration due to gravity at earth,
i.e, $\left(\because g_m=\frac{g_c}{6}\right)$
Squaring Eq. (iii) and putting this value,
$$
\begin{aligned}
&1=\frac{l_e}{l_m} \times \frac{g_e / 6}{g_e}=\frac{l_e}{l_m} \times \frac{1}{6} \\
&\frac{l_e}{6 l_m}=1 \Rightarrow 6 l_m=l_e \\
&l_m=\frac{1}{6} l_e=\frac{1}{6} \times 1=\frac{1}{6} m \\
&l_m=\frac{1}{6} m
\end{aligned}
$$
On the surface of the moon, the time period
$$
\begin{aligned}
&\left(T_m\right)=2 \pi \sqrt{\frac{l_m}{g_m}} \Rightarrow T_m^2=4 \pi^2\left(\frac{l_m}{g_m}\right) \\
&\left(\frac{T_e}{T_m}\right)^2=\frac{4 \pi^2}{4 \pi^2}\left(\frac{l_e}{g_e}\right) \times\left(\frac{g_m}{l_m}\right)
\end{aligned}
$$
Dividing eq. (i) from (ii),
$$
\left(\frac{T_e}{T_m}\right)=\sqrt{\frac{l_e}{l_m} \times \frac{g_m}{g_e}}
$$
$T_e=T_m=2 \mathrm{sec}$ to maintain the second's pendulum time period.
$$
1=\sqrt{\frac{l_e}{l_m} \times \frac{g_m}{g_e}}
$$
But the acceleration due to gravity at moon is $1 / 6$ of the acceleration due to gravity at earth,
i.e, $\left(\because g_m=\frac{g_e}{6}\right)$
Squaring Eq. (iii) and putting this value,
$$
\begin{aligned}
&1=\frac{l_e}{l_m} \times \frac{g_e / 6}{g_e}=\frac{l_e}{l_m} \times \frac{1}{6} \\
&\frac{l_e}{6 l_m}=1 \Rightarrow 6 l_m=l_e \\
&l_m=\frac{1}{6} l_e=\frac{1}{6} \times 1=\frac{1}{6} m \\
&l_m=\frac{1}{6} m
\end{aligned}
$$
For a simple pendulum , $T=2 \pi \sqrt{\frac{l}{g}}$
where, $l=$ length of the pendulum and $g=$ acceleration due to gravity on surface of the earth.
$$
\begin{aligned}
&T_e=2 \pi \sqrt{\frac{l_e}{g_e}} \\
&T_e^2=4 \pi^2\left(\frac{l_e}{g_e}\right)
\end{aligned}
$$
On the surface of the moon, the time period
$$
\begin{aligned}
&\left(T_m\right)=2 \pi \sqrt{\frac{l_m}{g_m}} \Rightarrow T_m^2=4 \pi^2\left(\frac{l_m}{g_m}\right) \\
&\left(\frac{T_e}{T_m}\right)^2=\frac{4 \pi^2}{4 \pi^2}\left(\frac{l_e}{g_e}\right) \times\left(\frac{g_m}{l_m}\right)
\end{aligned}
$$
Dividing eq. (i) from (ii),
$$
\left(\frac{T_e}{T_m}\right)=\sqrt{\frac{l_e}{l_m} \times \frac{g_m}{g_e}}
$$
$T_e=T_m=2$ sec to maintain the second's pendulum time period.
$$
1=\sqrt{\frac{l_e}{l_m} \times \frac{g_m}{g_e}}
$$
But the acceleration due to gravity at moon is $1 / 6$ of the acceleration due to gravity at earth,
i.e, $\left(\because g_m=\frac{g_c}{6}\right)$
Squaring Eq. (iii) and putting this value,
$$
\begin{aligned}
&1=\frac{l_e}{l_m} \times \frac{g_e / 6}{g_e}=\frac{l_e}{l_m} \times \frac{1}{6} \\
&\frac{l_e}{6 l_m}=1 \Rightarrow 6 l_m=l_e \\
&l_m=\frac{1}{6} l_e=\frac{1}{6} \times 1=\frac{1}{6} m \\
&l_m=\frac{1}{6} m
\end{aligned}
$$
On the surface of the moon, the time period
$$
\begin{aligned}
&\left(T_m\right)=2 \pi \sqrt{\frac{l_m}{g_m}} \Rightarrow T_m^2=4 \pi^2\left(\frac{l_m}{g_m}\right) \\
&\left(\frac{T_e}{T_m}\right)^2=\frac{4 \pi^2}{4 \pi^2}\left(\frac{l_e}{g_e}\right) \times\left(\frac{g_m}{l_m}\right)
\end{aligned}
$$
Dividing eq. (i) from (ii),
$$
\left(\frac{T_e}{T_m}\right)=\sqrt{\frac{l_e}{l_m} \times \frac{g_m}{g_e}}
$$
$T_e=T_m=2 \mathrm{sec}$ to maintain the second's pendulum time period.
$$
1=\sqrt{\frac{l_e}{l_m} \times \frac{g_m}{g_e}}
$$
But the acceleration due to gravity at moon is $1 / 6$ of the acceleration due to gravity at earth,
i.e, $\left(\because g_m=\frac{g_e}{6}\right)$
Squaring Eq. (iii) and putting this value,
$$
\begin{aligned}
&1=\frac{l_e}{l_m} \times \frac{g_e / 6}{g_e}=\frac{l_e}{l_m} \times \frac{1}{6} \\
&\frac{l_e}{6 l_m}=1 \Rightarrow 6 l_m=l_e \\
&l_m=\frac{1}{6} l_e=\frac{1}{6} \times 1=\frac{1}{6} m \\
&l_m=\frac{1}{6} m
\end{aligned}
$$
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