
The bob in a simple pendulum of length $l$ is released at $t = 0$ from the position of small angular displacement ${\theta _0}$. Linear displacement of the bob at any time $t$ from the mean position is given by:
(A) $l{\theta _0}\cos \sqrt {\dfrac{g}{l}} t$
(B) $l\sqrt {\dfrac{g}{l}} t\cos {\theta _0}$
(C) $lg\sin {\theta _0}$
(D) $l{\theta _0}\sin \sqrt {\dfrac{g}{l}} t$
Answer
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Hint The linear displacement is the product of the amplitude and the cosine of the angular frequency with respect to the time. And the amplitude of the linear displacement is the product of the length of the pendulum and the angular displacement.
Useful formula
The linear displacement at any time is given by,
$x = a\cos \omega t$
Where, $x$ is the linear displacement, $a$ is the amplitude, $\omega $ is the angular frequency and $t$ is the time taken.
Complete step by step solution
Given that,
The length of the pendulum is $l$
The angular displacement of the pendulum is ${\theta _0}$
And the time is $t$.
Now, the amplitude $a$ is given by the product of the length and the angular displacement, then
$a = l{\theta _0}\,......................\left( 1 \right)$
Now, the angular frequency is given by the square root of the acceleration due to gravity divided by the length, then the angular frequency is written as,
$\omega = \sqrt {\dfrac{g}{l}} \,....................\left( 2 \right)$
Now,
The linear displacement at any time is given by,
$x = a\cos \omega t$
By substituting the amplitude of the pendulum of equation (1) and the angular frequency of the pendulum of equation (2) in the above equation of the linear displacement, then the above equation is written as,
$x = l{\theta _0}\cos \sqrt {\dfrac{g}{l}} t$
Thus, the above equation shows the linear displacement of the bob at any time.
Hence, the option (A) is the correct answer.
Note: From the final answer, the linear displacement of the pendulum or the bob is depends on the length of the pendulum or bob and the angular displacement of the pendulum or bob and the angular frequency of the pendulum or bob and the time.
Useful formula
The linear displacement at any time is given by,
$x = a\cos \omega t$
Where, $x$ is the linear displacement, $a$ is the amplitude, $\omega $ is the angular frequency and $t$ is the time taken.
Complete step by step solution
Given that,
The length of the pendulum is $l$
The angular displacement of the pendulum is ${\theta _0}$
And the time is $t$.
Now, the amplitude $a$ is given by the product of the length and the angular displacement, then
$a = l{\theta _0}\,......................\left( 1 \right)$
Now, the angular frequency is given by the square root of the acceleration due to gravity divided by the length, then the angular frequency is written as,
$\omega = \sqrt {\dfrac{g}{l}} \,....................\left( 2 \right)$
Now,
The linear displacement at any time is given by,
$x = a\cos \omega t$
By substituting the amplitude of the pendulum of equation (1) and the angular frequency of the pendulum of equation (2) in the above equation of the linear displacement, then the above equation is written as,
$x = l{\theta _0}\cos \sqrt {\dfrac{g}{l}} t$
Thus, the above equation shows the linear displacement of the bob at any time.
Hence, the option (A) is the correct answer.
Note: From the final answer, the linear displacement of the pendulum or the bob is depends on the length of the pendulum or bob and the angular displacement of the pendulum or bob and the angular frequency of the pendulum or bob and the time.
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