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Question

In a gravity free space, a particle is connected with four springs as shown below. If the particle is pulled down gently and left, then find the period of oscillation of the particle.
(Given:θ=45, β=30)


A
T=2πm4k
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B
T=2πm6k
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C
T=2π3m4k
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D
T=2πm2k
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Solution

The correct option is D T=2πm2k
Suppose the particle is displaced by a distance x downward. Then, the upper two springs will be elongated and lower springs will be compressed. The force will act as shown in the FBD given below:


Let y be the elongation in the upper springs. The net restoring force by the upper springs is


F=F1cos(90θ)+F2cos(90θ)
By symmetry, we can say that, F1=F2=F (say)
F=2Fsinθ .....(1)

Let the lower springs be compressed by y each.
The net restoring force by the two lower springs is given by
F′′=F3cos(90β)+F4cos(90β)
By symmetry, we can say that F3=F4=f (say)
F′′=2fsinβ ......(2)

From (1) and (2), we get
Fnet=2Fsinθ+2fsinβ
Fnet=(2kysinθ+4kysinβ)
From the first diagram, we can deduce that
Elongation in upper spring y=xcos(90θ)=xsinθ
Elongation in lower spring y=xcos(90β)=xsinβ
Therefore,
Fnet=(2kxsin2θ+4kxsin2β)
From the data given in the question,
Fnet=[2kx12+4kx14]
Fnet=2kxa=2kxm
Comparing this with a=ω2x, we get
ω=2kmT=2πm2k
Thus, option (d) is the correct answer.

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