The period of a simple pendulum is given by the series in Eq. 17-25. (a) For what value of m is the second term of the series equal to 0.02? (b) What is the value of the third term in the series at this amplitude?

To tackle your question about the period of a simple pendulum and the series given in Eq. 17-25, let's break it down into two parts: finding the value of \( m \) for which the second term equals 0.02, and then determining the value of the third term at that amplitude.

Understanding the Series for the Pendulum Period

The period \( T \) of a simple pendulum can be expressed as a series expansion, particularly when considering larger amplitudes. The general form of the series is:

• First term: \( T_0 = 2\pi \sqrt{\frac{L}{g}} \)
• Second term: \( T_1 = \frac{1}{2} T_0 \left( \frac{m}{2} \right) \)
• Third term: \( T_2 = \frac{1}{4} T_0 \left( \frac{m^2}{4} \right) \)

Here, \( L \) is the length of the pendulum, \( g \) is the acceleration due to gravity, and \( m \) is a parameter related to the amplitude of the pendulum's swing.

Finding the Value of m

To find the value of \( m \) such that the second term of the series equals 0.02, we set up the equation:

Equation: \( T_1 = \frac{1}{2} T_0 \left( \frac{m}{2} \right) = 0.02 \)

Rearranging gives:

Step 1: Multiply both sides by 2:

\( T_0 \left( \frac{m}{2} \right) = 0.04 \)

Step 2: Now, isolate \( m \):

\( m = \frac{0.04 \cdot 2}{T_0} \)

Substituting \( T_0 = 2\pi \sqrt{\frac{L}{g}} \) into the equation:

\( m = \frac{0.08}{2\pi \sqrt{\frac{L}{g}}} \)

Now, you would need the specific values of \( L \) and \( g \) to calculate \( m \). For example, if \( L = 1 \) meter and \( g = 9.81 \, \text{m/s}^2 \), then:

\( T_0 = 2\pi \sqrt{\frac{1}{9.81}} \approx 0.63 \, \text{s} \)

Substituting this back gives:

\( m \approx \frac{0.08}{0.63} \approx 0.127 \)

Calculating the Third Term

Now that we have \( m \), we can find the value of the third term \( T_2 \) at this amplitude:

Equation: \( T_2 = \frac{1}{4} T_0 \left( \frac{m^2}{4} \right) \)

Substituting \( m \) into the equation:

\( T_2 = \frac{1}{4} T_0 \left( \frac{(0.127)^2}{4} \right) \)

Step 1: Calculate \( m^2 \):

\( m^2 \approx 0.0161 \)

Step 2: Substitute back into the equation:

\( T_2 = \frac{1}{4} T_0 \left( \frac{0.0161}{4} \right) \)

Continuing with our earlier calculation of \( T_0 \):

\( T_2 = \frac{1}{4} \cdot 0.63 \cdot \frac{0.0161}{4} \approx 0.00025 \, \text{s} \)

Final Thoughts

In summary, the value of \( m \) that makes the second term equal to 0.02 can be calculated using the length of the pendulum and gravitational acceleration. Once \( m \) is determined, you can find the third term of the series, which provides insight into how the period of the pendulum changes with amplitude. This approach not only helps in understanding the dynamics of pendulums but also illustrates the importance of series expansions in physics.