Question

Two simple pendulums of length Im and 4m respectively are both given small
displacement in the same direction at the same instant. They will be again in phase
after the shorter pendulum has completed number of oscillations equal to

Answer 1

Answer:

Given:

Pendulum of length 1 m and 4 m have been used. They are given small displacement in the same direction at the same Instant.

To find:

Number of Oscillations of shorter pendulum after which both the pendulums will be in phase .

Concept:

We can follow this rule for solving this type of questions ; Let the number of Oscillations be denoted as "n":

Let n1 be for lathe pendulum , n2 for smaller pendulum

The pendulums will be in relation like :

$(n1)(t1) = (n2)(t2)$

$= > (n1)(2\pi \sqrt{ \dfrac{l1}{g} } ) = (n2)(2\pi \sqrt{ \dfrac{l2}{g} } )$

Cancelling similar terms :

$= > (n1)( \sqrt{l1} ) = (n2)( \sqrt{l2} )$

$= > \dfrac{(n1)}{(n2)} = \dfrac{ \sqrt{l2} }{ \sqrt{l1} }$

$= > \dfrac{(n1)}{(n2)} = \dfrac{ \sqrt{1} }{ \sqrt{4} }$

$= > \dfrac{(n1)}{(n2)} = \dfrac{1}{2}$

For every 2 Oscillations of smaller pendulum , both the Pendulums will be in phase .

Answer 2

Answer :

• For every two oscillations of smal pendulum, both the pendulums will be in the phase .

Explanation :

As we know that :

$\longrightarrow \sf{T \: = \: 2 \pi \sqrt{\dfrac{l}{g}}}\: \: \: \: \: ...(1)$

Where,

• T is time period
• l is length of pendulum
• g is acceleration due to gravity

And, we know that

$\longrightarrow \sf{Time \: \times \: No. \: of \: oscillations \: = \: constant}$

Let, Number of oscillations be n

So,

$\longrightarrow \sf{n_1 \: \times \: T_1 \: = \: n_2 \: \times \: T_2} \\ \\ \longrightarrow \sf{n_1 \: \times \: 2 \pi \sqrt{\dfrac{l_1}{g}} \: = \: n_2 \: \times \: 2 \pi \sqrt{\dfrac{l_2}{g}}} \\ \\ \longrightarrow \sf{n_1 \: \times \: \sqrt{l_1} \: = \: n_2 \: \times \: \sqrt{l_2}} \\ \\ \longrightarrow \sf{\dfrac{n_1}{n_2} \: = \: \dfrac{\sqrt{l_1}}{\sqrt{l_2}}} \\ \\ \longrightarrow \sf{\dfrac{n_1}{n_2} \: = \: \dfrac{\sqrt{1}}{\sqrt{4}}} \\ \\ \longrightarrow \sf{\dfrac{n_1}{n_2} \: = \: \dfrac{1}{2}}$