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meerkat18

It is given that the centre of mass of both the parakeets is 17 cm below the pivot.

We can therefore assume that all the mass is centred at 17 cm below the pivot and hence it can now be seen as a simple pendulum of length 17cm (0.17 m).

For a simple pendulum, time period of oscillation is T = 2*pi*sqrt(length/g) , where g is the acceleration due to gravity, g=9.8m/s^2

=> frequency = 1/T = 1/(2*pi*sqrt(length/g)) = 1/(2*pi*sqrt(0.17/9.8)) = 11.8422 Hz

Therefore they swing with a frequency of 11.8422 Hz.

 

nuhulawal20

The two parakeets sitting on the swing with a combined center of mass 17.0 cm below the pivot swings with a frequency of 1.21Hz.

Given the data in the question

Length of pendulum; [tex]L = 17.0cm = 0.17m[/tex]

Frequency; [tex]f = \ ?[/tex]

Using the simple pendulum equation of time:

[tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]-------let this be equation 1

Where T is the period of time, L is the length of the pendulum and g is the acceleration due to gravity( [tex]9.8m/s^2[/tex]).

Also formula for Frequency:

[tex]f = \frac{1}{T}[/tex] ------------let this be equation 2

Lets substitute equation 1 into equation 2

[tex]f = \frac{1}{2\pi \sqrt{\frac{L}{g} }}\\\\f = \frac{1}{2\pi } \sqrt{\frac{g}{L} }[/tex]

We substitute or values into the equation

[tex]f = \frac{1}{2\pi } \sqrt{\frac{9.8m/s^2}{0.17m} } \\\\f = \frac{1}{2\pi } \sqrt{57.647s^{-2}} \\\\f = \frac{1}{2\pi } *7.59256s^{-1}}\\\\f = 1.21 Hz[/tex]

Therefore, the two parakeets sitting on the swing with a combined center of mass 17.0 cm below the pivot swings with a frequency of 1.21Hz.

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