What is the period of a $1.0-m$-long pendulum on (a) the earth and (b) Venus?

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

$T=2\pi \sqrt{\frac{L}{g}}$

where $g=9.80\mathrm{m}/{\mathrm{s}}^2$is that the acceleration of gravity.

Note that the amount of a 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 is: $L=1.0m$

In part (a), we are asked to determine the period of the pendulum on the earth.

In part (b), we are asked to determine the period of the pendulum on Venus.

(a)

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

$T=2\pi \sqrt{\frac{L}{g_{\mathrm{E}}}}$

where $g_{\mathrm{E}}=9.80\mathrm{m}/{\mathrm{s}}^2$ on the earth.

Substitute numerical values:

$T=2\pi \sqrt{\frac{1.0\mathrm{m}}{9.80\mathrm{m}/{\mathrm{s}}^2}}$

localid="1650018123655" $=2.0s$

(b)

The period of a pendulum of Venus is given by an equation $(*)$:

localid="1650018132275" $T=2\pi \sqrt{\frac{L}{g_{\mathrm{V}}}}$

where $g_{\mathrm{V}}=8.87\mathrm{m}/{\mathrm{s}}^2$on Venus.

Substitute numerical values:

localid="1650018136647" $T=2\pi \sqrt{\frac{1.0\mathrm{m}}{8.8\mathrm{m}/{\mathrm{s}}^2}}$

$=2.1s$