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Grade 12th passMechanics

Answer the question please

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Samriddhi Sinha

11 Years agoGrade 12th pass

1 Answer

Sandeep Pathak

11 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 are $x_1^0$ and $x_2^0$ respectively.

Then tension in lower string is $T_2 = 3kx_2^0=mg$ .
Tension in upper string is $T_1 = 2T_2=6kx_2^0 = 2kx_1^0\Rightarrow x_1^0 = 3x_2^0$ .

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

Again, use $T_1 = 2T_2\Rightarrow x_1 = 3x_2$

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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