91影视

Given in the question that the Orangutan swing like simple pendulum with length of the simple pendulum $L=0.90m$ and at an angle$胃=20掳$

The length of the Orangutan arms is Apply the equation to find the period T of a simple pendulum of length L. It is given by,

$T=2\mathrm{蟺}\sqrt{\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{g}$}\right.}$

The angle of swinging of the orangutan is $胃=20掳$. The Orangutan is modeled as a pendulum.

Where g is the acceleration due to gravity. Notice that the period of a simple pendulum depends only on its length and on the magnitude of the gravitational constant. It does not depend on the mass of the object hanging at its end or the amplitude of vibration.

The average speed of particle is equal to the ratio of the total distance it travels to the total interval during which it travels that distance:

$v\left(avg\right)=\frac{d}{鈭�t}$

$T=2\mathrm{蟺}\sqrt{\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{g}$}\right.}$

Substitute the known values of L and $胃$,

$T=2\mathrm{蟺}\sqrt{\frac{0.90\mathrm{m}}{9.8\mathrm{m}/\mathrm{s}}}$
$=1.904s$

The time taken by the orangutan to move from one end to another end will be half of the time period.

Therefore,

$$\begin{gathered} t=\frac{T}{2} \\ =\frac{1.904s}{2} \\ =0.952s \end{gathered}$$

Therefore, the speed of the pendulum will be approximately,

$$\begin{gathered} v=\frac{d}{t} \\ =\frac{2L\sin 20掳}{0.952s} \\ =\frac{2(0.9m)\sin 20掳}{0.952s} \end{gathered}$$

$$\begin{gathered} =0.647m/s \\ =0.65m/s \end{gathered}$$