S1 and S2 are two sources of light which produce individually disturbance at point P given by E1 - 3 sin- t and E2 - 4cos- t- Assuming E1 and E2 to be along the same line- the resultant of their super position is

A vector diagram for amplitude of waves at point P is shown in the figure. Given, E_1 = 3 sinω t And, E_2 = 4 sin nbsp; ( ω t + π2 ) So, phase difference, ...

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Byju's Answer

Standard XII

Physics

Periodic Motion

S1 and S2 are...

Question

$S_{1}$ and $S_{2}$ are two sources of light which produce individually disturbance at point $P$ given by ${\to E}_{1} = 3 sin ω t$ and ${\to E}_{2} = 4 cos ω t$ . Assuming $_{1}$ and $_{2}$ to be along the same line, the resultant of their super position is

5 sin (ω t + 53^{\circ})

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5 sin (ω t + 37^{\circ})

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5 sin (ω t + 45^{\circ})

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5 sin (ω t + 30^{\circ})

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Solution

The correct option is A $5 sin (ω t + 53^{\circ})$
A vector diagram for amplitude of waves at point P is shown in the figure.

Given,
${\to E}_{1} = 3 sin ω t$
And, ${\to E}_{2} = 4 sin (ω t + \frac{π}{2})$

So, phase difference, $ϕ = \frac{π}{2}$

Therefore, resultant amplitude

$E = \sqrt{E_{1}^{2} + E_{2}^{2} + 2 E_{1} E_{2} cos ϕ}$

$= \sqrt{3^{2} + 4^{2} + (2 \times 3 \times 4 \times cos \frac{π}{2}}) = 5$

Also, the phase difference of the resultant disturbance is given by

$tan θ = \frac{E_{2} sin ϕ}{E_{1} + E_{2} cos ϕ}$

$= \frac{4 sin \frac{π}{2}}{3 + 4 cos \frac{π}{2}} = \frac{4}{3}$

$\Rightarrow θ = {tan}^{- 1} (\frac{4}{3}) = 53^{\circ}$

Hence, net disturbance, $E = 5 sin (ω t + 53^{\circ})$

Hence, option $(A)$ is correct.

Why this question?
This question give you ides that amplitudes of waves can be added vectorially if the waves are on the same line.

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S1 and S2 are two sources of light which produce individually disturbance at point P given by →E1=3sinωt and →E2=4cosωt. Assuming −→E1 and −→E2 to be along the same line, the resultant of their super position is

$S_{1}$ and $S_{2}$ are two sources of light which produce individually disturbance at point $P$ given by ${\to E}_{1} = 3 sin ω t$ and ${\to E}_{2} = 4 cos ω t$ . Assuming $_{1}$ and $_{2}$ to be along the same line, the resultant of their super position is