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Question

3 SHMs of equal amplitude A and equal time periods in the same direction combine. The phase of 2nd motion is 60 ahead of the first and phase of 3rd is 60 ahead of second. Find the amplitude of the resultant.

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A
A
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B
3/2A
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C
2A
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D
3A
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Solution

The correct option is C 2A


We are given that
x1=Asinωtx2=Asin(ωt+π3)x3=Asin(ωt+2π3)
so the equation of the resultant is
x1+x2+x3=A(sinωt+sin(ωt+π3)+sin(ωt+2π3))
adding the first term (sinωt) and third term (sin(ωt+2π3)) using trigonometric formula sinA+sinB=2sin(A+B2)cos(AB2), we have
x1+x2+x3=A(2sin(ωt+π3)cos(π3)+sin(ωt+π3))x1+x2+x3=A(2sin(ωt+π3)×12+sin(ωt+π3))x1+x2+x3=A(sin(ωt+π3)+sin(ωt+π3))x1+x2+x3=2Asin(ωt+π3)
Now we can see that the amplitude of the resultant SHM is 2A


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