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Question

A point mass is subjected to two simultaneous sinusoidal displacements along the x - direction: x1(t)=Asinωt and x2(t)=Asin(ωt+2π3). Adding a third sinusoidal displacement x3(t)=Bsin(ωt+ϕ) brings the mass to a complete rest. The values of B and ϕ are respectively:

A
2A and 3π4
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B
A and 4π3
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C
3A and 5π6
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D
A and π3
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Solution

The correct option is B A and 4π3
Given,
x1(t)=Asinωt
x2(t)=Asin(ωt+2π3)
x3(t)=Bsin(ωt+ϕ)
We know that, mass will be at rest when Fnet=0

From principle of superposition
Fnet=F1+F2+F3
When the mass is at rest,
F3=(F1+F2) .......(1)

Using Hooke's law, we can rewrite (1) as
x3=(x1+x2)
[since Fx in SHM]
Bsin(ωt+ϕ)=(Asinωt+Asin(ωt+2π3))

Using, sinC+sinD=2sin(C+D2)cos(CD2)
we get,

Bsin(ωt+ϕ)=2Asin(ωt+π3)cos(π3)
Bsin(ωt+ϕ)=Asin(ωt+π3)
We know that sin(π+θ)=sinθ
Bsin(ωt+ϕ)=Asin(ωt+4π3)

Therfore, Amplitude of the third sinusoidal displacement is A and Initial phase ϕ=4π3.

Thus, option (b) is the correct answer.


Alternate solution:
To bring the mass to complete rest, the net force acting on the particle should be zero. The sinusoidal displacement vector can be represented as an equilateral triangle.



Hence, from the phasor diagram , it can be seen that amplitude of the third displacement should be A and phase constant should be 4π3.
Hence, (b) is the required answer

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