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The resultant amplitude of a vibrating particle by the superposition of the two ways $y_{1} = a \sin (ω t + \frac{π}{3})$ and $y_{2} = a \sin ω t$ is

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$\sqrt{2} a$

$\sqrt{3} a$

answer is D.

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Detailed Solution

$y_{1} = a \sin (ω t + \frac{π}{3})$

$y_{2} = a \sin ω t$

$θ = ω t + \frac{π}{3} - ω t$

$θ = \frac{A}{3}$

R = $\sqrt{a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} \cos θ}$

= $\sqrt{a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} \cos \frac{π}{3}}$

R = $\sqrt{3} a$

so option : 4

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