Question

calcular la longitud de un péndulo que oscila a 10 Hz en santa fe de bogota, sabiendo que en esta ciudad la aceleracion de la gravedad es de 978 cm/s2.

Answer 1

Answer:

$L=2.48*10^{-3} m$

Explanation:

The period equation for a pendulum is given by:

$T=2\pi \sqrt{\frac{L}{g}}$

and we know that T = 1/f, where f is the frequency, so we will have:

$\frac{1}{f}=2\pi \sqrt{\frac{L}{g}}$

Now, we just need to solve this equation for L.

$\frac{1}{2\pi f}=\sqrt{\frac{L}{g}}$

$L=\frac{g}{(2\pi f)^{2}}$

• g is the gravity in Bogota (g=9.78 m/s^{2})
• f is 10 Hz
• L is the lenght of the pendulum

$L=\frac{9.78}{(2\pi*10)^{2}}$

$L=2.48*10^{-3} m$

I hope it helps you!