Question

the period of a pendulum is given by the equation

$$t = 2pi sqrt{\frac{l}{g}},$$

where ( l ) is the length of the string suspending the pendulum in meters, and ( g ) is the acceleration due to gravity in ( \text{m/s}^2 ).

which of the following domains provide a real - value period?

• ( g < 0 )
• ( g = 0 )
• ( g > 0 )
• ( g geq 0 )

done

Explanation:

Step1: Analyze the square root

For the square root $\sqrt{\frac{L}{g}}$ to be a real number, the expression inside the square root (the radicand) must be non - negative, and the denominator cannot be zero. Since $L$ (length of the string) is a positive quantity (you can't have a negative length), we need to consider the sign of $g$.
The radicand is $\frac{L}{g}$. Since $L>0$, for $\frac{L}{g}\geq0$, and also $g
eq0$ (because it is in the denominator), we need $g > 0$. If $g<0$, then $\frac{L}{g}<0$ and the square root is not a real number. If $g = 0$, then we have a division by zero, which is undefined. If $g\geq0$, when $g = 0$ it is undefined. So only when $g>0$, the expression $\sqrt{\frac{L}{g}}$ is a real number, and thus the period $T$ is a real number.

Answer:

$g>0$