Answer:
(a) Natural frequency = 0.99 rad/sec (b) 0.4086 rad/sec
Explanation:
We have given length of pendulum = 10 m
(a) Acceleration due to gravity [tex]=9.81m/sec^2[/tex]
Time period of pendulum is given by [tex]T=2\pi\sqrt{\frac{L}{g}}[/tex], L is length of pendulum and g is acceleration due to gravity
So [tex]T=2\pi\sqrt{\frac{L}{g}}=2\times 3.14\times \sqrt{\frac{10}{9.81}}=6.34sec[/tex]
Natural frequency is given by [tex]\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{6.34}=0.99rad/sec[/tex]
(b) In this case acceleration due to gravity [tex]g=1.67m/sec^2[/tex]
So time period [tex]T=2\pi\sqrt{\frac{L}{g}}=2\times 3.14\times \sqrt{\frac{10}{1.67}}=15.3674sec\[/tex]
Natural frequency [tex]\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{15.36}=0.4086rad/sec[/tex]
