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Answer the question please

Samriddhi Sinha , 10 Years ago
Grade 12th pass
anser 1 Answers
Sandeep Pathak

Last Activity: 10 Years ago

Since we are talking about small oscillations, we have to consider them about the equilibrium position. Let’s call upper spring, spring 1 and lower spring, spring 2. Also assume the equilibrium extensions of springs arex_1^0and x_2^0respectively.

Then tension in lower string isT_2 = 3kx_2^0=mg.
Tension in upper string isT_1 = 2T_2=6kx_2^0 = 2kx_1^0\Rightarrow x_1^0 = 3x_2^0.

Now assume that the two springs are extended by x_1 and x_2 respectively.Then the mass lowers by an amountx=2x_1+x_2. Let us calculate restoring force on mass
F=3k(x_2^0+x_2)-mg=3kx_2

Again, useT_1 = 2T_2\Rightarrow x_1 = 3x_2

Thus,
x=2x_1+x_2=7x_2\Rightarrow x_2 = \frac{x}{7}

Restoring force,
F=\frac{3kx}{7}\Rightarrow \omega = \sqrt{\frac{7m}{3k}}\Rightarrow n=3

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