Question

The given setup is in equilibrium springs were in natural length before force was applied. If I have to replace the above setup of 1 block and 2 springs by 1 block and 1 spring, What will be the spring constant of that new spring, such that the block is displaced by the same amount as previously.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Spring1 and spring 2 will have same expansion so they are in parallel.

2 parallel spring with spring constant $K_1$ & $K_2$ can be replaced with 1 spring of constant ($K_1+K_2$).

Now this new spring will be in series with the spring 3.

So for series equivalent is

$\frac{1}{K_1+K_2}+\frac{1}{K_3}=\frac{1}{K_{net}}$

$K_{net}=\frac{K_3(K_1+K_2)}{K_1+K_2+K_3}$

Alternate solution

$X_3=\frac{F}{K_2}$ .....(i) $K_3{X}_3=(K_1+K_2)X_1=F$ $\Rightarrow X_1=\frac{F}{K_1+K_2}$ ..........(ii)

$F=K_0{X}_{net}$

$X_0=\frac{F}{K_{onet}}$ .......(iii)

$X_0=X_1+X_3$

$\Rightarrow \frac{F}{K_{0net}}=\frac{F}{K_1+K_2}+\frac{F}{K_2}$

$\frac{1}{K_{net}}=\frac{1}{K_1+K_2}+\frac{1}{K_3}$

$K_{net}=\frac{K_3(K_1+K_2)}{K_1+K_2+K_3}$