Question

A simple pendulum 0.64m long has a period of 1.2seconds. Calculate the period of a similar pendulum 0.36m long in the same location.

Answer 1

The period of the second pendulum is 0.9 s

Explanation:

The period of a simple pendulum is given by the equation

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity at the location of the pendulum

For the first pendulum, we have

L = 0.64 m

T = 1.2 s

Therefore we can find the value of g at that location:

$g=(\frac{2\pi}{T})^2 L=(\frac{2\pi}{1.2})^2 (0.64)=17.5 m/s^2$

Now we can find the period of the second pendulum at the same location, which is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where we have

L = 0.36 m (length of the  second pendulum)

$g=17.5 m/s^2$

Substituting,

$T=2\pi \sqrt{\frac{0.36}{17.5}}=0.9 s$

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