Answer:
Explanation:
Because we assume the pendulum is a "mathematical pendulum" (neglecting the moment of inertia of the bob), we can find:
[tex]T=2\pi\sqrt{\frac{L}{g}} \rightarrow 4=2\pi\sqrt{\frac{L}{9.81}} \rightarrow \frac{4}{\pi^{2}}=\frac{L}{9.81} \rightarrow L \approx 3.97 m[/tex]
By using the [tex]y=A\sin(\omega t) \rightarrow v = \frac{dy}{dt}=\omega A \cos\omega t = \omega\sqrt{A^{2}-y^{2}}[/tex]
The mean position is the position when y = 0, so:
[tex]\omega = \frac{2\pi}{T}=\frac{2\pi}{4}=0.5\pi[/tex] rad/s
and [tex]v = \omega A \rightarrow A=\frac{40}{0.5\pi}=\frac{80}{\pi}[/tex] in centimeters (cm).
