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Classical Mechanics Level 4

The above system comprises of three ideal springs, and two thick rods (they don't look like a rod!).The Temperature of the above system is increased by $\displaystyle\Delta T$ . The coefficient of linear expansion of the material of the rods is $\displaystyle \alpha$ .

Calculate the energy stored in the spring with spring constant $\displaystyle 2K$ .

Details and Assumptions:
$\bullet$ The springs are initially relaxed.
$\bullet$ The rods expand only in the horizontal direction.
$\bullet$ Neglect the expansion of the springs due to the increase in temperature
$\bullet$ $\displaystyle L = 1m$
$\bullet$ $\displaystyle K = 200 N/m$
$\bullet$ $\displaystyle \alpha = 2\times 10^{-5} K^{-1}$
$\bullet$ $\displaystyle \Delta T = 20^o C$

The answer is 0.000005355.

1 solution

Jatin Yadav
Apr 12, 2014

Let the compression in springs be $x_{1}, x_{2}$ , and $x_{3}$ .

On balancing forces on rods,

$x_{1}=2x_{2}= 3x_{3}$ .

Hence, $x_{1} = 2x_{2}, x_{3} = \frac{2}{3} x_{2}$

Now, the elongation of rods is balanced by compression of springs, hence,

$x{1}+x_{2}+x_{3} = \frac{3}{2} L \alpha \triangle T$

Thus, $x_{2} = \frac{9L}{22} \alpha \triangle T$

Required energy stored = $\frac{1}{2} 2K x_{2}^2 = K \bigg(\frac{9L}{22} \alpha \triangle T\bigg)^2 \approx 0.000005355$

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The answer is 0.000005355.

1 solution

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