Question

What is the period of a simple pendulum 47 cm long (a) on the Earth, and ( b) when it is in a freely falling elevator?

Answer 1

Answer:

a)1.37 s

b)∞ ( Infinite)

Explanation:

Given that

L= 47 cm              ( 1 m =100 cm)

L= 0.47 m

a)

On the earth :

Acceleration due to gravity = g

We know that time period of the simple pendulum given as

$T=2\pi\sqrt{ \dfrac{L}{g_{{eff}}}$

Here

$g_{eff}= g$

Now by putting the values

$T=2\pi \times\sqrt{ \dfrac{0.47}{9.81}}$

T=1.37 s

b)

Free falling elevator :

When elevator is falling freely then

$g_{eff}= 0$            ( This is case of weightless motion)

Therefore

$T=2\pi\sqrt{ \dfrac{L}{0}$

T=∞  (Infinite)