(B) Given that the amplitude of each sinusoidal wave is the same.Let $a_1 = a_2 = a$.The phase difference is $\phi = 120^{\circ}$.The resultant amplit

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(B) Given that the amplitude of each sinusoidal wave is the same.Let $a_1 = a_2 = a$.The phase difference is $\phi = 120^{\circ}$.The resultant amplitude is $A = 2 \,mm$.The formula for the resultant amplitude of two interfering waves is given by:$A = \sqrt{a_1^2 + a_2^2 + 2 a_1 a_2 \cos \phi}$Substituting the given values:$2 = \sqrt{a^2 + a^2 + 2 a^2 \cos 120^{\circ}}$Since $\cos 120^{\circ} = -\frac{1}{2}$, we have:$2 = \sqrt{a^2 + a^2 + 2 a^2 (-0.5)}$$2 = \sqrt{2a^2 - a^2}$$2 = \sqrt{a^2}$$2 = a$Therefore, the amplitude of each wave is $2 \,mm$.

Class 11 Physics Waves and Sound Principle of superposition of waves

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Two identical sinusoidal waves are moving in the same direction along a stretched string and interfere with each other. The phase difference between them is $120^{\circ}$. The amplitudes of both the waves are the same. If the amplitude of the resultant wave due to interference is $2 \,mm$, the amplitude of each wave is:

TS EAMCET 2019 All TS EAMCET

A
$1 \,mm$
B
$2 \,mm$
C
$\sqrt{3} \,mm$
D
$2 \sqrt{3} \,mm$

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Waves and Sound

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Principle of superposition of waves

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