Question

The velocity of a particle is given by v = v₀ sin ωt, where v₀ is constant and ω = 2π/T. Find the average velocity in time Interval t = 0 to t = T/2​

Answer 1

Given :-

• v = v₀ sin ωt
• ω = 2π/T
• t = 0 to t = T/2

To find :-

• Find the average velocity.

Let's Do...!!

$\sf \: v =\dfrac{ ∫_{0} {}^{T/2} vdt}{ ∫_{0} {}^{T/2} dt} = \sf\dfrac{∫_{0} {}^{T/2}v_{0} \: sin \: wt.dt}{(\dfrac{t}{2})}$

$\sf \purple{:\implies} \dfrac{2 \: v_{0}}{T} \ ∫_{0} {}^{T/2} \: sin(wt)dt$

$\: \: \: \: \purple{: \implies}\sf\dfrac{2 \: v_{0}}{T}( \frac{ - cos \: wt}{w} {)}^{T/2} _{0}$

$\: \: \: \: \: \: \: \: \: \: \: \purple{: \implies} \sf\dfrac{2 \: v_{0}}{wt}(1 - cos \: \dfrac{wt}{2})$

$\: \: \: \: \: \: \: \: \: \purple{: \implies} \sf \: w = \dfrac{2\pi}{T} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{:\implies}\sf w \times\dfrac{T}{2} = \cancel\dfrac{2\pi}{T} \times \cancel\dfrac{2\pi}{T} = \pi$

$\: \: \: \: \: \: \: \: \: \: \: \purple{: \implies}\sf\dfrac{2 \: v_{0}}{wt} \: (1 - cos\pi) = \dfrac{2 \: v_{0}}{2\pi} {}^{(2)}$

★ Final answer is :

$\: \: \: \: \: \: \: \: \red⇝\boxed{\sf \: v = \sf\dfrac{2 \: v_{0}}{\pi}}$

Answer 2

☆Concept:

》 Average velocity is defined as Total displacement covered between the time given.

$\boxed{\bf{ v_{av} = \dfrac{\Delta x}{\Delta t} = \dfrac{dx}{dt}}}$

☆Solution:

We will apply here simple integration!

• $\rm v = v_0 sin\omega t$

• $\rm \dfrac{dx}{dt} = v_0 sin\omega t$

• $\rm dx = (v_0 sin\omega t)dt$

• $\rm \int dx = v_0 \int (sin\omega t)dt$

• $\rm \int dx = \dfrac{v_0 \int -cos\omega t}{\omega}$

•$\rm \Delta x = \dfrac{-v_0 \int _{0}^{T/2} cos\omega t}{\omega}$

•$\rm \Delta x = \dfrac{-v_0}{\omega}(cos\dfrac{2\pi}{T}.\dfrac{T}{2})-(cos\dfrac{2\pi}{T} 0)$

•$\rm \Delta x = \dfrac{-v_0}{\omega}(cos\pi - cos 0°)$

• $\rm \Delta x = \dfrac{-v_0}{\omega}(-1 - 1)$

• $\rm \Delta x = \dfrac{-v_0}{\omega}(-2)$

•$\rm \Delta x = \dfrac{2v_0}{\omega}$

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》$\Delta t = \dfrac{T}{2} - 0$

•$\rm v_{av} = \dfrac{\Delta x}{\Delta t}$

•$\rm v_{av} = \dfrac{2v_0}{\omega (T/2 -0)}$

•$\rm v_{av} = \dfrac{2v_0}{\dfrac{2\pi}{T}(T/2)}$

•$\bf v_{av} = \dfrac{2v_0}{\pi}$

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