Answer :
To determine the domain in which the period of a pendulum provides a real value, we analyze the given formula for the period of a pendulum:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
Let's examine each case:
1. Case \( g < 0 \):
- If \( g < 0 \), then the expression inside the square root, \( \frac{L}{g} \), becomes negative since \( L \) (the length of the string) is a positive value by definition.
- The square root of a negative number is not a real number. Hence, the period \( T \) will not be a real number.
- Therefore, the period is not real for \( g < 0 \).
2. Case \( g = 0 \):
- If \( g = 0 \), the denominator of the fraction \( \frac{L}{g} \) becomes zero.
- Division by zero is undefined. Consequently, \( T \) is not defined when \( g = 0 \).
- Hence, the period is undefined for \( g = 0 \).
3. Case \( g > 0 \):
- If \( g > 0 \), the expression inside the square root, \( \frac{L}{g} \), remains positive since both \( L \) and \( g \) are positive.
- The square root of a positive number is a real number. Therefore, \( T \) will be a real number.
- Thus, the period is real for \( g > 0 \).
4. Case \( g \geq 0 \):
- If \( g \ge 0 \), we include both cases \( g = 0 \) and \( g > 0 \).
- As analyzed earlier, the period \( T \) is not defined for \( g = 0 \) and real for \( g > 0 \).
- Since \( g = 0 \) makes the period undefined, \( g \ge 0 \) is not a valid domain for \( T \) to be real.
In conclusion:
- The period is not real for \( g < 0 \).
- The period is undefined for \( g = 0 \).
- The period is real for \( g > 0 \).
- The period is undefined at \( g = 0 \) but real for \( g > 0 \).
Thus, the domains that provide a real value period are:
- \( g < 0 \): False
- \( g = 0 \): False
- \( g > 0 \): True
- [tex]\( g \ge 0 \)[/tex]: False
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
Let's examine each case:
1. Case \( g < 0 \):
- If \( g < 0 \), then the expression inside the square root, \( \frac{L}{g} \), becomes negative since \( L \) (the length of the string) is a positive value by definition.
- The square root of a negative number is not a real number. Hence, the period \( T \) will not be a real number.
- Therefore, the period is not real for \( g < 0 \).
2. Case \( g = 0 \):
- If \( g = 0 \), the denominator of the fraction \( \frac{L}{g} \) becomes zero.
- Division by zero is undefined. Consequently, \( T \) is not defined when \( g = 0 \).
- Hence, the period is undefined for \( g = 0 \).
3. Case \( g > 0 \):
- If \( g > 0 \), the expression inside the square root, \( \frac{L}{g} \), remains positive since both \( L \) and \( g \) are positive.
- The square root of a positive number is a real number. Therefore, \( T \) will be a real number.
- Thus, the period is real for \( g > 0 \).
4. Case \( g \geq 0 \):
- If \( g \ge 0 \), we include both cases \( g = 0 \) and \( g > 0 \).
- As analyzed earlier, the period \( T \) is not defined for \( g = 0 \) and real for \( g > 0 \).
- Since \( g = 0 \) makes the period undefined, \( g \ge 0 \) is not a valid domain for \( T \) to be real.
In conclusion:
- The period is not real for \( g < 0 \).
- The period is undefined for \( g = 0 \).
- The period is real for \( g > 0 \).
- The period is undefined at \( g = 0 \) but real for \( g > 0 \).
Thus, the domains that provide a real value period are:
- \( g < 0 \): False
- \( g = 0 \): False
- \( g > 0 \): True
- [tex]\( g \ge 0 \)[/tex]: False
