Two parakeets sit on a swing with their combined center of mass 17.0 cm below the pivot. at what frequency (in hz) do they swing?

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.

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg