Three S.H.M's of equal amplitude A and equal time period in the same direction combine. The difference in phase between each pair is $60^{\circ}$ ahead of the other. The amplitude of the resultant oscillation is :

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Solution

The correct option is B
2a

Resultant of first two with phase difference, $(ϕ = 60^{\circ})$ will be $R_{1} = \sqrt{3 a}$ with $90^{\circ}$ diffrence in phase with the third. The resultant of third and

$R_{1}$ is

$R = \sqrt{(\sqrt{3} a)^{2} + a^{2}} = 2 a$

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Three S.H.M's of equal amplitude A and equal time period in the same direction combine. The difference in phase between each pair is 60∘ ahead of the other. The amplitude of the resultant oscillation is :

The correct option is B 2a Resultant of first two with phase difference, (ϕ=60∘) will be R1=√3a with 90∘ diffrence in phase with the third. The resultant of third and R1 is R=√(√3a)2+a2=2a

Three S.H.M's of equal amplitude A and equal time period in the same direction combine. The difference in phase between each pair is $60^{\circ}$ ahead of the other. The amplitude of the resultant oscillation is :

The correct option is B
2a

Resultant of first two with phase difference, $(ϕ = 60^{\circ})$ will be $R_{1} = \sqrt{3 a}$ with $90^{\circ}$ diffrence in phase with the third. The resultant of third and

$R_{1}$ is

$R = \sqrt{(\sqrt{3} a)^{2} + a^{2}} = 2 a$