Question

Two springs of force constants $300N/m$ (spring $A$) and $400N/m$ (spring $B$) are joined together in series. The combination is compressed by $8.75cm.$ The ratio of energy stored in $a$ and $B$ is $\frac{E_A}{E_B}.$
Then $\frac{E_A}{E_B}$ is equal to

Answer 1

For springs in series the load carried is same as the load applied throughout the series.

Hence $300x_a=400x_b$

Also $x_a+x_b=0.0875m$

By Solving the equations we get

$x_a=0.05m$

$x_b=0.0375m$

Energy in first spring , $E_A=0.5\times 300\times {0.05}^2=0.375J$

Energy in second spring, $E_B=0.5\times 400\times {0.0375}^2=0.281J$

Required ratio $\frac{E_A}{E_B}=\frac{0.375}{0.281}=1.33$