Answer:
If we double the length we will have:
[tex]f'=\frac{f}{\sqrt{2}}[/tex]
Explanation:
The equation of the angular frequency of a pendulum is given by:
[tex]\omega=\sqrt{\frac{g}{L}}[/tex]
Where:
By definition the period is:
[tex]T=\frac{2\pi}{\omega}[/tex]
And frequency is 1 over the frequency T:
[tex]f=\frac{1}{T}[/tex]
[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{L}}[/tex]
Now, if we double the length we will have:
[tex]f'=\frac{1}{2\pi}\sqrt{\frac{g}{2L}}[/tex]
[tex]f'=\frac{1}{2\pi\sqrt{2}}\sqrt{\frac{g}{L}}[/tex]
[tex]f'=\frac{f}{\sqrt{2}}[/tex]
Therefore, the new frequency is the old frequency over √2.
I hope it helps you!
