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

`L=2.48*10^(-3) m`

Step-by-step explanation:

The period equation for a pendulum is given by:

`T=2\pi \sqrt{(L)/(g)}`

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

`(1)/(f)=2\pi \sqrt{(L)/(g)}`

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

`(1)/(2\pi f)=\sqrt{(L)/(g)}`

`L=(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=(9.78)/((2\pi*10)^(2))`

`L=2.48*10^(-3) m`

I hope it helps you!