Question

A block of mass m is attached to two unstretched springs of spring constants k1 and k2 as shown in figure.The block is displaced towards right through x and is released. Find the speed of the block as it passed through the mean position.

Answer 1

Solution:- Assume that the speed of the block = v 

Displacement in the body = x 

Now for the potential energy , which is stored in the spring at the time of compression .

Potential Energy , P.E = 1/2k₁x₁² + 1/2k₂x₂²

In the mean position, by law of conservation of energy 

So, Potential energy changes to Kinetic energy of the block.

1/2mv² = 1/2k₁x₁² +1/2k₂x₂² 

mv² = x² (k₁ + k₂)

v² = x² (k₁ +k₂ ) /m

$v \: = \sqrt{ (\frac{k1 + k2}{m} )x}$

Hence the speed of the block when it passes through their mean position is 

$v = \sqrt{( \frac{k1 + k2}{m}) x}$

Hope it helps you !!

Answer 2

Explanation:

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