Question

A block of mass m is attached to four unstretched massless springs of spring constant k1 and k2 , as

shown in above figure. The block is displaced towards right through distance x and is released. Speed of block when displacement of block is x/2 from mean position​

Answer 1

The equivalent spring constant of the two springs each of spring constant $\sf{k_1}$ is equal to $\sf{2k_1,}$ since they're in parallel.

So we can replace them by a spring of spring constant $\sf{2k_1.}$

Similarly, the two springs each of spring constant $\sf{k_2}$ can be replaced by a spring of spring constant $\sf{2k_2.}$

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(0,0)(70,0){2}{\line(0,1){15}}\put(0,0){\line(1,0){70}}\put(30,0){\framebox(10,10){\sf{m}}}\multiput(0,5)(40,0){2}{\multiput(0,0)(22.5,0){2}{\line(1,0){7.5}}\multiput(0,0)(1.25,0){10}{\qbezier(7.5,0)(8.75,5)(10,0)\qbezier(10,0)(9.375,-5)(8.75,0)}\qbezier(20,0)(21.25,5)(22.5,0)}\multiput(0,0)(71.5,0){2}{\multiput(0,0)(0,2){8}{\qbezier(0,0)(-0.5,-0.5)(-1,-1)}}\multiput(2,0)(2,0){35}{\qbezier(0,0)(-0.5,-0.5)(-1,-1)}\put(12.5,10.5){ \sf{2k_1} }\put(52.5,10.5){ \sf{2k_2} }\end{picture}$

As the block is displaced through $\sf{x}$ and then released, it has potential energy due to elasticity of the springs only there.

$\sf{\longrightarrow U_1=\dfrac{1}{2}(2k_1)x^2+\dfrac{1}{2}(2k_2)x^2}$

$\sf{\longrightarrow U_1=(k_1+k_2)x^2}$

and,

$\sf{\longrightarrow K_1=0}$

After release, the block will have a velocity $\sf{v}$ at a distance $\sf{\dfrac{x}{2}}$ from mean position. So it has kinetic energy there.

$\sf{\longrightarrow K_2=\dfrac{1}{2}\,mv^2}$

And its potential energy will be,

$\sf{\longrightarrow U_2=\dfrac{1}{2}(2k_1)\left(\dfrac{x}{2}\right)^2+\dfrac{1}{2}(2k_2)\left(\dfrac{x}{2}\right)^2}$

$\sf{\longrightarrow U_2=\dfrac{1}{4}(k_1+k_2)x^2}$

By energy conservation,

$\sf{\longrightarrow K_1+U_1=K_2+U_2}$

$\sf{\longrightarrow (k_1+k_2)x^2=\dfrac{1}{2}\,mv^2+\dfrac{1}{4}(k_1+k_2)x^2}$

$\sf{\longrightarrow\dfrac{1}{2}\,mv^2=\dfrac{3}{4}(k_1+k_2)x^2}$

$\sf{\longrightarrow\underline{\underline{v=x\sqrt{\dfrac{3(k_1+k_2)}{2m}}}}}$