Understanding Pendulum Time Period and Gravity

This question asks us to compare the time periods of two pendulum clocks under different gravitational conditions. The time period of a simple pendulum depends on its length and the acceleration due to gravity, but not on the mass of the bob.

Key Formula for Simple Pendulum

The time 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.
• \(g\) is the acceleration due to gravity at the location of the pendulum.

Notice that the mass of the bob is not present in this formula, indicating that the time period is independent of the bob's mass.

Analyzing the First Pendulum Clock

For the first pendulum clock, let its length be \(L_1\), the mass of its bob be \(m_1\), and the gravitational acceleration be \(g_1\). Its time period is given as \(T\).

• Length, \(L_1 = L\)
• Mass, \(m_1 = m\)
• Gravitational acceleration, \(g_1 = g\)
• Time period, \(T_1 = T\)

Using the formula, we have:

$$T_1 = T = 2\pi \sqrt{\frac{L_1}{g_1}} = 2\pi \sqrt{\frac{L}{g}}$$

Analyzing the Second Pendulum Clock

For the second pendulum clock, its length is the same as the first, the mass of its bob is double, and the gravitational acceleration is half.

• Length, \(L_2 = L\) (same length as the first)
• Mass, \(m_2 = 2m\) (double the mass of the first)
• Gravitational acceleration, \(g_2 = g/2\) (half the value)
• Let its time period be \(T_2\)

Using the formula, the time period \(T_2\) is:

$$T_2 = 2\pi \sqrt{\frac{L_2}{g_2}}$$

Substitute the values \(L_2 = L\) and \(g_2 = g/2\):

$$T_2 = 2\pi \sqrt{\frac{L}{g/2}}$$

Calculating the Time Period of the Second Pendulum

Simplify the expression for \(T_2\):

$$T_2 = 2\pi \sqrt{\frac{L}{g/2}} = 2\pi \sqrt{\frac{L \times 2}{g}} = 2\pi \sqrt{\frac{2L}{g}}$$

We can separate the \(\sqrt{2}\) term:

$$T_2 = 2\pi \sqrt{2} \sqrt{\frac{L}{g}}$$

From the first pendulum, we know that \(T = 2\pi \sqrt{\frac{L}{g}}\). Substitute this into the equation for \(T_2\):

$$T_2 = \sqrt{2} \times \left(2\pi \sqrt{\frac{L}{g}}\right)$$

$$T_2 = \sqrt{2} T$$

Effect of Mass on Pendulum Time Period

It's important to note that while the mass of the bob for the second pendulum is double that of the first, this information is irrelevant to the time period calculation of an ideal simple pendulum. The formula \(T = 2\pi \sqrt{L/g}\) does not include mass.

Summary of Results

Comparing the time periods:

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

$$T_2 = 2\pi \sqrt{\frac{L}{g/2}} = 2\pi \sqrt{\frac{2L}{g}} = \sqrt{2} \times 2\pi \sqrt{\frac{L}{g}} = \sqrt{2} T$$

Thus, the time period of the second pendulum is \(\sqrt{2}\) times the time period of the first pendulum.

Revision Table: Pendulum Clock Concepts

Additional Information: Pendulum Clocks and Gravity

Pendulum clocks rely on the constant time period of a pendulum for accurate timekeeping. Because the time period depends on the local gravitational acceleration (\(g\)), a pendulum clock will run faster or slower if moved to a location with a different gravitational pull. For example:

• If \(g\) increases (e.g., moving from the equator towards the poles), the time period \(T\) decreases, and the clock runs faster.
• If \(g\) decreases (e.g., moving from the poles towards the equator, or going up a mountain), the time period \(T\) increases, and the clock runs slower.

The question describes lifting the clocks to different heights, which implies changes in the gravitational acceleration \(g\). Gravity decreases with altitude. The time period of a pendulum clock is directly proportional to the square root of its length and inversely proportional to the square root of the gravitational acceleration.

$$T \propto \sqrt{L}$$

$$T \propto \frac{1}{\sqrt{g}}$$

Understanding these relationships helps in predicting how a pendulum clock's timekeeping will change under varying conditions of length and gravity.