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If the period of oscillation of a simple pendulum is 4s, find its length. If the velocity of the bob
at the mean position is 40cms−1
, find its amplitude. (take gravity = 9.81ms−2

Sagot :

deriyan191293

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).

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