Question

simple pendulum has a period t on earth. if it were used on planet x, where the acceleration 198) due to gravity is 3 times what it is on earth, its period would be:

Answer 1

The simple pendulum will have a period which is
`(1)/(√(3))` times the period on Earth on plant x.

The formula for period T of a simple pendulum is:

`\[ T = 2\pi \sqrt{(L)/(g)} \]`

where:

T - time

L - length of the pendulum,

g - acceleration due to gravity

If the acceleration due to gravity on planet X is three times that on Earth
`\(g_{\text{X}} = 3g_{\text{Earth}}\)`, and the length of the pendulum remains the same, we can compare the periods on Earth
`\(T_{\text{Earth}}\)` and on planet x
`\(T_{\text{x}}\)`:

`\[ T_{\text{x}} = 2\pi \sqrt{\frac{L}{g_{\text{x}}}} \]`

Substituting
`\(g_{\text{x}} = 3g_{\text{Earth}}\)`:

`\[ T_{\text{x}} = 2\pi \sqrt{\frac{L}{3g_{\text{Earth}}}} \]`

Comparing with the equation for time on Earth,

`\[ \frac{T_{\text{x}}}{T_{\text{Earth}}} = \frac{2\pi \sqrt{\frac{L}{3g_{\text{Earth}}}}}{2\pi \sqrt{\frac{L}{g_{\text{Earth}}}}} \]`

`\[ \frac{T_{\text{x}}}{T_{\text{Earth}}} = \sqrt{\frac{g_{\text{Earth}}}{3g_{\text{Earth}}}} \]`

`\[ \frac{T_{\text{x}}}{T_{\text{Earth}}} = \sqrt{(1)/(3)} \]`

`\[ T_{\text{x}} = (1)/(√(3) )} \cdot T_{\text{Earth}} \]`

So, the period of the simple pendulum on planet x would be
`(1)/(√(3))` times the period on Earth.