91Ó°ÊÓ

The period $T$of an easy pendulum of length $L$is given by,

localid="1650024523013" $T=2\mathrm{Ï€}\sqrt{\frac{L}{g}}$

where $g$is that the acceleration of gravity.

Note thatthe amount ofa straightforward pendulum depends only on its length and the magnitude of itsconstant of gravitation.

Itdoesn't rely upon the mass ofthe item hanging fromthe sting or the amplitude of vibration.

The length of the pendulum on earth is: $L_{\mathrm{E}}=2.0\mathrm{m}$.

We are asked to determine the length of a pendulum whose period on the moon equals the period of a $2.0m$-long pendulum on earth.

The period of the pendulum on the planet is found from Equation(*) :

$T=2\mathrm{Ï€}\sqrt{\frac{L_{\mathrm{E}}}{g_{\mathrm{E}}}}$

where $L_{\mathrm{E}}$is the length of the earth pendulum and $g_{\mathrm{E}}$is the acceleration due to gravity on Earth.

Similarly, the amount of the pendulum on the moon is:

localid="1650024551328" $T=2\mathrm{Ï€}\sqrt{\frac{L_{\mathrm{M}}}{g_{\mathrm{M}}}}$

We know that the acceleration due to gravity $g_{\mathrm{M}}$on the moon is $1/6$times the acceleration due to gravity $g_{\mathrm{E}}$on the world. So,

localid="1650024554821" $T=2\mathrm{Ï€}\sqrt{\frac{L_{\mathrm{M}}}{\frac{g_{\mathrm{E}}}{6}}}$

localid="1650024559612" $=\sqrt{6}\left(2\mathrm{Ï€}\sqrt{\frac{L_M}{g_{\mathrm{E}}}}\right)$

Setting Equations (1) and (2) equal, we obtain:

localid="1650024563364" $2\mathrm{Ï€}\sqrt{\frac{L_{\mathrm{E}}}{g_{\mathrm{E}}}}=\sqrt{6}\left(2\mathrm{Ï€}\sqrt{\frac{L_{\mathrm{M}}}{g_{\mathrm{E}}}}\right)$

localid="1650024567330" $\sqrt{L_{\mathrm{E}}}=\sqrt{6}\sqrt{L_{\mathrm{M}}}$

Square both sides:

localid="1650024570866" $L_{\mathrm{E}}=6L_{\mathrm{M}}$

Solve forlocalid="1650024584044" $L_{\mathrm{M}}$:

localid="1650024579598" $L_{\mathrm{M}}=\frac{L_{\mathrm{E}}}{6}$

localid="1650024588287" $=\frac{2.0\mathrm{m}}{6}$

localid="1650024575669" $=0.3\mathrm{m}$