Question

Two simple pendulums of length 5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed - oscillations.

• A. 2
• B. 5
• C. 1
• D. 3

Answer 1

Answer: (A)

2

Answer 2

Frequency of the pendulum
$\upsilon_{ l = 5 } = \frac{1}{2 \pi} \sqrt{ \frac{ g}{ 5}}$

$\upsilon_{ l = 2 0 } = \frac{1}{ 2 \pi} \sqrt{ \frac{ g}{ 20 }}$

$\therefore \frac{ \upsilon_{ l = 5 }}{ \upsilon_{ l = 20 }} = \sqrt{ \frac{ 20}{ 5} } = 2$
$\Rightarrow \upsilon_{ l = 5 } = 2 \upsilon_{ l = 20 }$
As shorter length pendulum has frequency double the larger length pendulum. Therefore shorter pendulum should complete 2 oscillations before they will be again in phase.

So, the correct option is (A): 2