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The period of a pendulum is given by the equation:
[tex]\[ T = 2 \pi \sqrt{\frac{L}{g}} \][/tex]

where [tex][tex]$L$[/tex][/tex] is the length of the string suspending the pendulum in meters, and [tex][tex]$g$[/tex][/tex] is the acceleration due to gravity in [tex][tex]$m/s^2$[/tex][/tex].

Which of the following domains provide a real value period?

A. [tex][tex]$g\ \textless \ 0$[/tex][/tex]
B. [tex][tex]$g=0$[/tex][/tex]
C. [tex][tex]$g\ \textgreater \ 0$[/tex][/tex]
D. [tex][tex]$g \geq 0$[/tex][/tex]

Sagot :

GinnyAnswer
To determine which domains provide a real value for the period [tex]\( T \)[/tex] of a pendulum, we need to analyze the given equation:

[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 pendulum.
- [tex]\( g \)[/tex] is the acceleration due to gravity.

For the period [tex]\( T \)[/tex] to be real, the expression inside the square root must be non-negative, because the square root of a negative number is not a real number. Thus, the term [tex]\(\frac{L}{g}\)[/tex] must be non-negative.

Since [tex]\( L \)[/tex] (the length of the pendulum) is always positive, the denominator [tex]\( g \)[/tex] must be positive to ensure that [tex]\(\frac{L}{g} \geq 0\)[/tex].

Let's analyze each case:

1. [tex]\( g < 0 \)[/tex]:
- If [tex]\( g \)[/tex] is negative, then [tex]\(\frac{L}{g}\)[/tex] would be negative.
- A negative value inside the square root results in an imaginary number, not a real number.
- Therefore, [tex]\( T \)[/tex] is not real in this domain.

2. [tex]\( g = 0 \)[/tex]:
- If [tex]\( g \)[/tex] is zero, we would be attempting to divide by zero in [tex]\(\frac{L}{g}\)[/tex], which is undefined in mathematics.
- An undefined expression cannot yield a real value.
- Therefore, [tex]\( T \)[/tex] is not real in this domain.

3. [tex]\( g > 0 \)[/tex]:
- If [tex]\( g \)[/tex] is positive, then [tex]\(\frac{L}{g}\)[/tex] would be positive since [tex]\( L \)[/tex] is also positive.
- The square root of a positive number is a real number.
- Therefore, [tex]\( T \)[/tex] is real in this domain.

4. [tex]\( g \geq 0 \)[/tex]:
- If [tex]\( g \)[/tex] is zero, the expression [tex]\(\frac{L}{0}\)[/tex] becomes undefined, as discussed earlier.
- If [tex]\( g \)[/tex] is positive, then [tex]\(\frac{L}{g}\)[/tex] is positive and results in a real number.
- Since zero cannot be included, this domain as a whole does not guarantee [tex]\( T \)[/tex] being real for all values.
- Therefore, [tex]\( T \)[/tex] is not real in this domain as a whole.

In conclusion, the domain that provides a real value for the period [tex]\( T \)[/tex] of the pendulum is:

[tex]\( g > 0 \)[/tex].
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