A particle is subjected to two SHMs of displacement x1-5sin- t-30- and x2-3sin- t-120- respectively- Find out the amplitude of the resulting SHM-

Given, nbsp;x_1=5sin(ω t+30^∘) and x_2=3sin(ω t+120^∘) Also, A_1=5, A_2=3 and phase difference ϕ=90^∘ Thus, using formula we have A_net=√(A_1^2+A_2^2+2× A_1 ...

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Standard XII

Physics

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A particle is...

Question

A particle is subjected to two SHMs of displacement $x_{1} = 5 sin (ω t + 30^{\circ})$ and $x_{2} = 3 sin (ω t + 120^{\circ})$ respectively. Find out the amplitude of the resulting SHM.

\sqrt{8}

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\sqrt{34}

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\sqrt{2}

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\sqrt{15}

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Solution

The correct option is B $\sqrt{34}$
Given, $x_{1} = 5 sin (ω t + 30^{\circ})$ and $x_{2} = 3 sin (ω t + 120^{\circ})$
Also, $A_{1} = 5, A_{2} = 3$ and phase difference $ϕ = 90^{\circ}$

Thus, using formula we have
$A_{n e t} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 \times A_{1} \times A_{2} cos ϕ}$
$\Rightarrow A_{n e t} = \sqrt{5^{2} + 3^{2} + 2 \times 5 \times 3 \times cos 90^{\circ}}$
$\Rightarrow A_{n e t} = \sqrt{25 + 9} = \sqrt{34}$

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A particle is subjected to two SHMs of displacement x1=5sin(ωt+30∘) and x2=3sin(ωt+120∘) respectively. Find out the amplitude of the resulting SHM.

The correct option is B √34Given, x1=5sin(ωt+30∘) and x2=3sin(ωt+120∘) Also, A1=5,A2=3 and phase difference ϕ=90∘ Thus, using formula we have Anet=√A21+A22+2×A1×A2cosϕ ⇒ Anet=√52+32+2×5×3×cos90∘ ⇒ Anet=√25+9=√34

A particle is subjected to two SHMs of displacement $x_{1} = 5 sin (ω t + 30^{\circ})$ and $x_{2} = 3 sin (ω t + 120^{\circ})$ respectively. Find out the amplitude of the resulting SHM.