To determine the domains that provide a real value for the period [tex]\(T\)[/tex] of a pendulum, let us analyze the given equation carefully:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
where:
- [tex]\(T\)[/tex] is the period of the pendulum
- [tex]\(L\)[/tex] is the length of the string in meters
- [tex]\(g\)[/tex] is the acceleration due to gravity in [tex]\( \text{m/s}^2 \)[/tex]
### Analysis by Case:
1. Case [tex]\( g < 0 \)[/tex]
- When [tex]\( g \)[/tex] is negative, the term [tex]\( \frac{L}{g} \)[/tex] becomes negative.
- The square root of a negative number is not a real number (it becomes an imaginary number).
- Therefore, for [tex]\( g < 0 \)[/tex], the period [tex]\( T \)[/tex] is not a real value.
2. Case [tex]\( g = 0 \)[/tex]
- When [tex]\( g = 0 \)[/tex], the term [tex]\( \frac{L}{g} \)[/tex] involves division by zero, which is undefined in mathematics.
- Hence, for [tex]\( g = 0 \)[/tex], the period [tex]\( T \)[/tex] is undefined.
3. Case [tex]\( g > 0 \)[/tex]
- When [tex]\( g \)[/tex] is positive, the term [tex]\( \frac{L}{g} \)[/tex] is positive.
- The square root of a positive number is real.
- Therefore, for [tex]\( g > 0 \)[/tex], the period [tex]\( T \)[/tex] is a real value.
4. Case [tex]\( g \geq 0 \)[/tex]
- This domain includes both [tex]\( g > 0 \)[/tex] and [tex]\( g = 0 \)[/tex].
- As analyzed, for [tex]\( g = 0 \)[/tex] the period [tex]\( T \)[/tex] is undefined.
- For [tex]\( g > 0 \)[/tex], the period [tex]\( T \)[/tex] is real.
- Thus, the domain [tex]\( g \geq 0 \)[/tex] does not uniformly provide real values (it includes a case where the value is undefined).
### Conclusion:
The only domain that guarantees the period [tex]\( T \)[/tex] of the pendulum to be a real value is:
[tex]\[ g > 0 \][/tex]
Thus, the correct domain which provides a real value for the period is:
[tex]\[ g > 0 \][/tex]
