
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 how many oscillations?
A) 5
B) 2
C) 1
D) 3
Answer
571.8k+ views
Hint: We need to analyze and understand the relation between the length of the pendulum strings and its dependence on the number of oscillations completed in a certain interval. We need to consider the time period of oscillation in these cases.
Complete step-by-step solution
We are given two pendulums of different lengths which are set to the same oscillations by some horizontal displacement. We know that the length of a pendulum determines the time period of oscillation of the pendulum. We know that the time period of oscillation of a pendulum is directly proportional to the square root of the length of the pendulum string.
It is given as –
\[\begin{align}
& T\propto \sqrt{l} \\
& \therefore T=2\pi \sqrt{\dfrac{l}{g}} \\
\end{align}\]
We know that the two given pendulums are set to the same displacement. As a result of this, the number of oscillations in a particular time interval or conversely, the time taken for a single oscillation depends only on the length of the pendulum.
Now, let us find the time period of the two pendulums of length 5 cm and 20 cm each.
\[\begin{align}
& {{T}_{1}}=2\pi \sqrt{\dfrac{0.05}{g}} \\
& \text{and,} \\
& {{T}_{2}}=2\pi \sqrt{\dfrac{0.20}{g}}=4\pi \sqrt{\dfrac{0.05}{g}} \\
\end{align}\]
Now, we can relate the time periods from the above equations as –
\[{{T}_{2}}=2{{T}_{1}}\]
From this we understand that the time period of oscillation of the second pendulum of length 20 cm takes twice as much time as that of the first pendulum of length 5 cm.
\[\begin{align}
& {{T}_{2}}=2{{T}_{1}} \\
& \therefore n=2 \\
\end{align}\]
We can see that the first pendulum of smaller length will undergo two oscillations by the time it coincides with the oscillation of the larger pendulum.
The correct answer is option B.
Note: The lengths of the pendulum strings are the only factor that can vary the time period of oscillation of two systems of pendulums irrespective of any other factors including the amplitude of the oscillation, i.e., the displacement from the mean position.
Complete step-by-step solution
We are given two pendulums of different lengths which are set to the same oscillations by some horizontal displacement. We know that the length of a pendulum determines the time period of oscillation of the pendulum. We know that the time period of oscillation of a pendulum is directly proportional to the square root of the length of the pendulum string.
It is given as –
\[\begin{align}
& T\propto \sqrt{l} \\
& \therefore T=2\pi \sqrt{\dfrac{l}{g}} \\
\end{align}\]
We know that the two given pendulums are set to the same displacement. As a result of this, the number of oscillations in a particular time interval or conversely, the time taken for a single oscillation depends only on the length of the pendulum.
Now, let us find the time period of the two pendulums of length 5 cm and 20 cm each.
\[\begin{align}
& {{T}_{1}}=2\pi \sqrt{\dfrac{0.05}{g}} \\
& \text{and,} \\
& {{T}_{2}}=2\pi \sqrt{\dfrac{0.20}{g}}=4\pi \sqrt{\dfrac{0.05}{g}} \\
\end{align}\]
Now, we can relate the time periods from the above equations as –
\[{{T}_{2}}=2{{T}_{1}}\]
From this we understand that the time period of oscillation of the second pendulum of length 20 cm takes twice as much time as that of the first pendulum of length 5 cm.
\[\begin{align}
& {{T}_{2}}=2{{T}_{1}} \\
& \therefore n=2 \\
\end{align}\]
We can see that the first pendulum of smaller length will undergo two oscillations by the time it coincides with the oscillation of the larger pendulum.
The correct answer is option B.
Note: The lengths of the pendulum strings are the only factor that can vary the time period of oscillation of two systems of pendulums irrespective of any other factors including the amplitude of the oscillation, i.e., the displacement from the mean position.
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