Dukell24idn
Answered

Get personalized and accurate responses to your questions with IDNLearn.com. Get accurate and comprehensive answers from our network of experienced professionals.

The period of a pendulum is given by the equation

[tex] T=2 \pi \sqrt{\frac{L}{g}} [/tex]

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

Which of the following domains provide a real value period?

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

Sagot :

GinnyAnsweridn
To determine which domains provide a real value for the period [tex]\( T \)[/tex] of a pendulum, we start with the given equation:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]

Here:
- [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]\( \text{m}/\text{s}^2 \)[/tex].

We need to analyze the expression inside the square root, [tex]\(\frac{L}{g}\)[/tex], to ensure that it yields a real number.

1. Domain [tex]\( g < 0 \)[/tex]:

For [tex]\( g < 0 \)[/tex], the expression [tex]\(\frac{L}{g}\)[/tex] becomes negative because [tex]\( L \)[/tex] is always positive (as it is a length). The square root of a negative number results in an imaginary number, which means the period [tex]\( T \)[/tex] would not be real.

Therefore, [tex]\( g < 0 \)[/tex] is invalid.

2. Domain [tex]\( g = 0 \)[/tex]:

For [tex]\( g = 0 \)[/tex], the expression [tex]\(\frac{L}{g}\)[/tex] becomes undefined because division by zero is not allowed in mathematics. Since this would result in an undefined operation, the period [tex]\( T \)[/tex] cannot be determined and is thus not real.

Therefore, [tex]\( g = 0 \)[/tex] is invalid.

3. Domain [tex]\( g > 0 \)[/tex]:

For [tex]\( g > 0 \)[/tex], the expression [tex]\(\frac{L}{g}\)[/tex] is positive because both [tex]\( L \)[/tex] and [tex]\( g \)[/tex] are positive. Taking the square root of a positive number results in a real number, which means the period [tex]\( T \)[/tex] will be real.

Therefore, [tex]\( g > 0 \)[/tex] is valid.

4. Domain [tex]\( g \geq 0 \)[/tex]:

For [tex]\( g \geq 0 \)[/tex], this includes both [tex]\( g > 0 \)[/tex] and [tex]\( g = 0 \)[/tex]. As analyzed above, [tex]\( g > 0 \)[/tex] is valid while [tex]\( g = 0 \)[/tex] is invalid because it leads to an undefined operation. Since this domain includes an invalid case, the entire domain [tex]\( g \geq 0 \)[/tex] cannot ensure a real value for the period.

Therefore, [tex]\( g \geq 0 \)[/tex] is invalid.

In summary, the valid domain that ensures a real value for the period [tex]\( T \)[/tex] of the pendulum is:
[tex]\[ g > 0 \][/tex]
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.