Question

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

• A. $\frac{4}{3}$
• B. $\frac{16}{9}$
• C. $\frac{3}{4}$
• D. $\frac{9}{16}$

Answer 1

The correct option is A $\frac{4}{3}$

It is given that,

Displacement of spring A is, $d_a=kx=300x_a$

Displacement of spring B is, $d_b=400x_b$

Since, $x_a+x_b=8.75cm=0.0875m............\left(1\right)$

When the springs are connected in series the load carried is same as the load applied throughout the series.

So, $300x_a=400x_b..........\left(2\right)$

Solving equations (1) and (2)

$x_a=0.05m$

$x_b=0.0375m$

Energy in first spring,

$E_a=\frac{1}{2}\times 300\times {0.05}^2...........\left(3\right)$

$E_b=\frac{1}{2}\times 400\times {0.0375}^2.......\left(4\right)$

Dividing equations (3) and (4)

$\frac{E_a}{E_b}=\frac{\frac{1}{2}\times 300\times {0.05}^2}{\frac{1}{2}\times 400\times {0.0375}^2}$

$\frac{E_a}{E_b}=\frac{4}{3}$

Ratio of energy stored in A and B is $4:3$