Questions: 2. How long must a pendulum be on the Moon, where g=1.6 N / kg, to have a period of 2.0 s ? 3. On a certain planet, the period of a 0.75-m-long pendulum is 1.8 s. What is g for this planet?

Solution Steps

The period \( T \) of a simple pendulum is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

To find the length \( L \) of the pendulum, rearrange the formula: \[ L = \frac{T^2 \cdot g}{4\pi^2} \]

Substitute \( T = 2.0 \, \mathrm{s} \) and \( g = 1.6 \, \mathrm{N/kg} \) into the formula: \[ L = \frac{(2.0)^2 \cdot 1.6}{4\pi^2} \]

Perform the calculation to find \( L \): \[ L = \frac{4 \cdot 1.6}{4\pi^2} = \frac{6.4}{39.4784} \approx 0.162 \, \mathrm{m} \]

Use the same formula for the period of a pendulum: \[ T = 2\pi \sqrt{\frac{L}{g}} \]

Rearrange the formula to solve for \( g \): \[ g = \frac{4\pi^2 \cdot L}{T^2} \]

Substitute \( L = 0.75 \, \mathrm{m} \) and \( T = 1.8 \, \mathrm{s} \) into the formula: \[ g = \frac{4\pi^2 \cdot 0.75}{(1.8)^2} \]

Perform the calculation to find \( g \): \[ g = \frac{29.6088 \cdot 0.75}{3.24} \approx 6.85 \, \mathrm{N/kg} \]

Final Answer