Two simple harmonic motions are represented by equations- y1-4sin 10t- and y2-5cos 10t- What is the phase difference between their velocities-

Given that, y_1=4sin (10t+ϕ) y_2=5cos 10t The velocity is given by nbsp; v_1 = ddt(4sin (10t+ϕ)) ⇒ v_1 =40cos (10t+ϕ) and nbsp; v_2 = ddt(5cos 10t) ⇒ v_2 = ...

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Standard XII

Physics

Phase Difference

Two simple ha...

Question

Two simple harmonic motions are represented by equations, $y_{1} = 4 sin (10 t + ϕ)$ and $y_{2} = 5 cos 10 t .$ What is the phase difference between their velocities?

ϕ

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- ϕ

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ϕ + \frac{π}{2}

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ϕ - \frac{π}{2}

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Solution

The correct option is D $ϕ - \frac{π}{2}$
Given that,
$y_{1} = 4 sin (10 t + ϕ)$
$y_{2} = 5 cos 10 t$
The velocity is given by
$v_{1} = \frac{d}{d t} (4 sin (10 t + ϕ))$
$\Rightarrow v_{1} = 40 cos (10 t + ϕ)$
and
$v_{2} = \frac{d}{d t} (5 cos 10 t)$
$\Rightarrow v_{2} = - 50 sin 10 t$

The phase difference between $v_{1}$ and $v_{2}$ is
$Δ ϕ = (10 t + ϕ) - (10 t + \frac{π}{2})$
$\Rightarrow Δ ϕ = ϕ - \frac{π}{2}$

Hence, the correct option is (d)

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Two simple harmonic motions are represented by equations, y1=4sin(10t+ϕ) and y2=5cos10t. What is the phase difference between their velocities?

The correct option is D ϕ−π2Given that, y1=4sin(10t+ϕ) y2=5cos10t The velocity is given by v1=ddt(4sin(10t+ϕ)) ⇒v1=40cos(10t+ϕ) and v2=ddt(5cos10t) ⇒v2=−50sin10t The phase difference between v1 and v2 is Δϕ=(10t+ϕ)−(10t+π2) ⇒Δϕ=ϕ−π2 Hence, the correct option is (d)

Two simple harmonic motions are represented by equations, $y_{1} = 4 sin (10 t + ϕ)$ and $y_{2} = 5 cos 10 t .$ What is the phase difference between their velocities?