Question

If a pendulum is 10m long, (a) what is the natural frequency and the period of vibration on the earth, where the free-fall acceleration is 9.81 m/s^2 and (b) what is the natural frequency and the period of vibration on the moon, where the free-fall acceleration is 1.67 m/s^2?

Answer 1

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 $=9.81m/sec^2$

Time period of pendulum is given by $T=2\pi\sqrt{\frac{L}{g}}$, L is length of pendulum and g is acceleration due to gravity

So $T=2\pi\sqrt{\frac{L}{g}}=2\times 3.14\times \sqrt{\frac{10}{9.81}}=6.34sec$

Natural frequency is given by $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{6.34}=0.99rad/sec$

(b) In this case acceleration due to gravity $g=1.67m/sec^2$

So time period $T=2\pi\sqrt{\frac{L}{g}}=2\times 3.14\times \sqrt{\frac{10}{1.67}}=15.3674sec$

Natural frequency $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{15.36}=0.4086rad/sec$