
Two simple pendulums of lengths 1 m and 16 m respectively are both given small displacements in the same direction at the same instant. They will again be in phase after the shorter pendulum has completed n oscillations where n is?
A. $1\dfrac{1}{3}$
B. $\dfrac{1}{4}$
C. 4
D. 5
Answer
161.1k+ views
Hint: In this question, we have given two pendulums and we have to find the oscillations when the shorter pendulum comes in the same phase. For this, we use the relation between the pendulum and the number of oscillations that it completes in a certain period of time and then calculate the number of oscillations.
Formula Used:
We know time period is defined by
$T=2\pi \sqrt{\dfrac{l}{g}} \\ $
And $w=\dfrac{2\pi }{T} \\ $
So $w=\sqrt{\dfrac{g}{l}}$
Where $w$ is the angular frequency, g is the gravity downward and L is the length of the pendulum.
Complete step by step solution:
Given length of two pendulums are $1m$ and $16m$ respectively. As per the pendulum, first we will find the angular frequency of the simple harmonic motion of the pendulum is given as:
$w=\sqrt{\dfrac{g}{L}}$
So, for the first pendulum, the angular frequency is:
${{w}_{1}}=\sqrt{\dfrac{g}{{{L}_{1}}}} $
And ${{w}_{1}}=\sqrt{\dfrac{g}{1}}$
Then ${{w}_{1}}={{w}_{0}}$
Now, angular frequency for the second pendulum is :
${{w}_{2}}=\sqrt{\dfrac{g}{{{L}_{2}}}}$
$\Rightarrow {{w}_{2}}=\sqrt{\dfrac{g}{16}}$
Then ${{w}_{2}}=\dfrac{1}{4}{{w}_{0}}$
Now the relative frequency of two pendulum:
${{w}_{1}}-{{w}_{2}}=\dfrac{3}{4}{{w}_{0}}$
In order to come in the same phase time taken by it is given by :
$t=\dfrac{\theta }{w} \\ $
$\Rightarrow t=\dfrac{2\pi }{\dfrac{3}{4}{{w}_{0}}} \\ $
That is $t=\dfrac{4}{3}\left( \dfrac{2\pi }{{{w}_{0}}} \right) \\ $
So, after completing $\dfrac{4}{3}$ or $1\dfrac{1}{3}$ phase, the shorter pendulum will be back in the same phase.
Thus, option A is the correct answer.
Note: The velocity of a pendulum changes with time but at a regular interval which is determined with the help of angular frequency. Students must have the knowledge of how to use the formula in a particular question so that they will be able to solve the question easily.
Formula Used:
We know time period is defined by
$T=2\pi \sqrt{\dfrac{l}{g}} \\ $
And $w=\dfrac{2\pi }{T} \\ $
So $w=\sqrt{\dfrac{g}{l}}$
Where $w$ is the angular frequency, g is the gravity downward and L is the length of the pendulum.
Complete step by step solution:
Given length of two pendulums are $1m$ and $16m$ respectively. As per the pendulum, first we will find the angular frequency of the simple harmonic motion of the pendulum is given as:
$w=\sqrt{\dfrac{g}{L}}$
So, for the first pendulum, the angular frequency is:
${{w}_{1}}=\sqrt{\dfrac{g}{{{L}_{1}}}} $
And ${{w}_{1}}=\sqrt{\dfrac{g}{1}}$
Then ${{w}_{1}}={{w}_{0}}$
Now, angular frequency for the second pendulum is :
${{w}_{2}}=\sqrt{\dfrac{g}{{{L}_{2}}}}$
$\Rightarrow {{w}_{2}}=\sqrt{\dfrac{g}{16}}$
Then ${{w}_{2}}=\dfrac{1}{4}{{w}_{0}}$
Now the relative frequency of two pendulum:
${{w}_{1}}-{{w}_{2}}=\dfrac{3}{4}{{w}_{0}}$
In order to come in the same phase time taken by it is given by :
$t=\dfrac{\theta }{w} \\ $
$\Rightarrow t=\dfrac{2\pi }{\dfrac{3}{4}{{w}_{0}}} \\ $
That is $t=\dfrac{4}{3}\left( \dfrac{2\pi }{{{w}_{0}}} \right) \\ $
So, after completing $\dfrac{4}{3}$ or $1\dfrac{1}{3}$ phase, the shorter pendulum will be back in the same phase.
Thus, option A is the correct answer.
Note: The velocity of a pendulum changes with time but at a regular interval which is determined with the help of angular frequency. Students must have the knowledge of how to use the formula in a particular question so that they will be able to solve the question easily.
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