Time period of a particle executing SHM is $8\sec $. At $t = 0$, it is at the mean position. The ratio of distance covered by the particle in the $1^{st}$ second to the $2^{nd}$ second is:
A) $\dfrac{1}{{\sqrt 2 - 1}}$
B) $\sqrt 2 $
C) $\dfrac{1}{{\sqrt 2 }}$
D) $\sqrt 2 + 1$
Answer
259.2k+ views
Hint: Recall that the simple harmonic motion is defined as the type of motion in which a restoring force acts. This restoring force is directly proportional to the displacement that a body experiences from its mean position. The direction of the restoring force is towards the mean position. In the simple harmonic motion, the path of the particle will always be in a straight line.
Complete step by step solution:
The equation of the simple harmonic motion for a particle is written as
$x = A\sin \omega t$---(i)
Where $\omega $ is the angular frequency
And t is the time taken by the particle
Also it is known that $\omega = \dfrac{{2\pi }}{T}$
At $t = 0\sec $the particle is at its mean position.
Given $T = 8\sec $
When $t = 1\sec $
${x_1} = A\sin (\dfrac{{2\pi }}{T}t)$
$\Rightarrow {x_1} = A\sin \dfrac{{2\pi }}{8}$
$\Rightarrow {x_1} = A\sin \dfrac{\pi }{4}$
Substitute the value of $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
The position of the particle at the 1st second is
$ \Rightarrow {x_1} = \dfrac{A}{{\sqrt 2 }}$
Similarly when $t = 2\sec $
${x_2} = A\sin (\dfrac{{2\pi }}{T}t)$
$\Rightarrow {x_2} = A\sin (\dfrac{{2\pi }}{8} \times 2)$
$\Rightarrow {x_2} = A\sin \dfrac{\pi }{2}$
Substitute the value of $\sin \dfrac{\pi }{2} = 1$
The position of the particle at the 2nd second is
$ \Rightarrow {x_2} = A$
The ratio of the distance travelled by the particle is given by
$\dfrac{{{x_1}}}{{{x_2} - {x_1}}} = \dfrac{{\dfrac{A}{{\sqrt 2 }}}}{{A - \dfrac{A}{{\sqrt 2 }}}}$
$ \Rightarrow \dfrac{A}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 A - A}}$
$ \Rightarrow \dfrac{A}{{A(\sqrt 2 - 1)}}$
$ \Rightarrow \dfrac{1}{{\sqrt 2 - 1}}$
Option A is the right answer.
Note: It is important to note that the simple harmonic motion is a type of oscillatory motion in which the acceleration of the particle is directly proportional to the displacement. Also all simple harmonic motions are oscillatory but all oscillatory motions are not simple harmonic. This is because oscillatory motion repeats itself after a fixed period of time. The mean position from which the particle started moving is in a stable equilibrium. This means that at the mean position the net force acting on the system will be zero.
Complete step by step solution:
The equation of the simple harmonic motion for a particle is written as
$x = A\sin \omega t$---(i)
Where $\omega $ is the angular frequency
And t is the time taken by the particle
Also it is known that $\omega = \dfrac{{2\pi }}{T}$
At $t = 0\sec $the particle is at its mean position.
Given $T = 8\sec $
When $t = 1\sec $
${x_1} = A\sin (\dfrac{{2\pi }}{T}t)$
$\Rightarrow {x_1} = A\sin \dfrac{{2\pi }}{8}$
$\Rightarrow {x_1} = A\sin \dfrac{\pi }{4}$
Substitute the value of $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
The position of the particle at the 1st second is
$ \Rightarrow {x_1} = \dfrac{A}{{\sqrt 2 }}$
Similarly when $t = 2\sec $
${x_2} = A\sin (\dfrac{{2\pi }}{T}t)$
$\Rightarrow {x_2} = A\sin (\dfrac{{2\pi }}{8} \times 2)$
$\Rightarrow {x_2} = A\sin \dfrac{\pi }{2}$
Substitute the value of $\sin \dfrac{\pi }{2} = 1$
The position of the particle at the 2nd second is
$ \Rightarrow {x_2} = A$
The ratio of the distance travelled by the particle is given by
$\dfrac{{{x_1}}}{{{x_2} - {x_1}}} = \dfrac{{\dfrac{A}{{\sqrt 2 }}}}{{A - \dfrac{A}{{\sqrt 2 }}}}$
$ \Rightarrow \dfrac{A}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 A - A}}$
$ \Rightarrow \dfrac{A}{{A(\sqrt 2 - 1)}}$
$ \Rightarrow \dfrac{1}{{\sqrt 2 - 1}}$
Option A is the right answer.
Note: It is important to note that the simple harmonic motion is a type of oscillatory motion in which the acceleration of the particle is directly proportional to the displacement. Also all simple harmonic motions are oscillatory but all oscillatory motions are not simple harmonic. This is because oscillatory motion repeats itself after a fixed period of time. The mean position from which the particle started moving is in a stable equilibrium. This means that at the mean position the net force acting on the system will be zero.
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