Question

If the time period of the oscillating body is \[t\] and amplitude is \[A\] . Then find the average velocity in SHM from the position of extreme point to the half of the amplitude \[\dfrac{A}{2}\] .

Answer 1

Hint:First of all, we will find the distance covered by the oscillating body from the extreme point to the position half the amplitude. Again, we will find the time required to cover this distance. After that we will find the average velocity by dividing the displacement by time.

Complete step by step solution:
In the given question, we are supplied the following data:
There is an oscillating body whose time period is given as \[t\] .
The amplitude of the vibrating body is given as \[A\] .
We are asked to find the average velocity in SHM from the position of extreme point to the half of the amplitude \[\dfrac{A}{2}\] .
To begin with, we will try to find the distance covered by the oscillating body to go from the extreme point to the half of the amplitude. After that we will find the total time taken by the body to complete one oscillation. After that, we will find the time taken by the body to cover the distance from the extreme point to the half of the amplitude. As we know that the velocity is defined as the displacement per unit time.
Let us proceed to solve the problem.
As we are given that the oscillating body goes from the extreme point to the half of the amplitude. So, let us calculate the distance in doing so.
Since, the total distance is \[A\] , so the distance from the extreme position to the half of the amplitude is \[\dfrac{A}{2}\] .
The total time period of an oscillating body is \[\dfrac{{2\pi }}{\omega }\] .
The duration of time to cover the distance from the extreme point to the half of the amplitude is:
$\dfrac{{\left( {\dfrac{{2\pi }}{\omega }} \right)}}{4} \\
\Rightarrow\dfrac{{2\pi }}{\omega } \times \dfrac{1}{4} \\
\Rightarrow \dfrac{\pi }{{2\omega }} \\$
The duration of time required to cover the distance from the extreme point to the half of the amplitude cover is found to be \[\dfrac{\pi }{{2\omega }}\] .
Now, we know:
\[{v_{{\text{avg}}}} = \dfrac{d}{t}\] …… (1)
Where,
\[{v_{{\text{avg}}}}\] indicates the average velocity of the oscillating body.
\[d\] indicates the distance covered by the body from the extreme position to the position half the amplitude.
\[t\] indicates the duration of time taken.
Now, we substitute the required values in the equation (1) and we get:
${v_{{\text{avg}}}} = \dfrac{d}{t} \\
\Rightarrow {v_{{\text{avg}}}} = \dfrac{{\dfrac{A}{2}}}{{\dfrac{\pi }{{2\omega }}}} \\
\Rightarrow {v_{{\text{avg}}}} = \dfrac{A}{2} \times \dfrac{{2\omega }}{\pi } \\
\Rightarrow {v_{{\text{avg}}}} = \dfrac{{A\omega }}{\pi } \\$

Hence, the average velocity is found to be \[\dfrac{{A\omega }}{\pi }\].

Note:While solving this problem always remember that the total distance covered by the oscillating body to complete one oscillation is \[4A\] . In the forward distance the distance covered by the body is \[2A\] while in the backward distance the distance covered is also the same i.e. \[2A\] . Velocity is maximum at mean position and hence the kinetic energy is also maximum at the mean position but the potential energy is minimum.