Question

Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?

Answer 1

`L=6.21m`

Step-by-step explanation:

For the simple pendulum problem we need to remember that:

`(d^(2)\theta)/(dt^(2))+(g)/(L)sin(\theta)=0`,

where
`\theta` is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

`\omega^(2)=(g)/(L)`,

where
`\omega` is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

`\omega=(2\pi)/(T)`,

where T is the oscillation period. Now, we can easily solve for L:

`((2\pi)/(T))^(2)=(g)/(L)\\\\L=g((T)/(2\pi))^(2)\\\\L=9.8((5)/(2\pi))^(2)\\\\L=6.21m`