A mass M is suspended by two springs A and B of force constants k1 and k2 respectively as shown in the diagram. The total stretch of springs in equilibrium is:

Question

A mass M is suspended by two springs A and B of force constants k1 and k2 respectively as shown in the diagram. The total stretch of springs in equilibrium is: <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-lkblzj5pkvmcof5y.jpg" /> (A) (M g/k1 + k2) (B) (2 Mg/(k1 + k2)) (C) (M g/2 (k1 + k2)) (D) (M g (k1 + k2)/k1 k2)

Tardigrade Admin · Accepted Answer

In series combination of spring, equivalent spring constant, (1/keq)=(1/k1)+(1/k2) ⇒ keq=(k1 k2/k1 + k2) At equilibrium, mg=keqx ⇒ x=(m g (k1 + k2)/k1 k2)

Q. A mass $M$ is suspended by two springs $A$ and $B$ of force constants $k_{1}$ and $k_{2}$ respectively as shown in the diagram. The total stretch of springs in equilibrium is:

$\frac{M g}{k _{1} + k _{2}}$

$\frac{2 M g}{( k _{1} + k _{2} )}$

$\frac{M g}{2 ( k _{1} + k _{2} )}$

$\frac{M g ( k _{1} + k _{2} )}{k _{1} k _{2}}$

In series combination of spring, equivalent spring constant,
$\frac{1}{k _{e q}} = \frac{1}{k _{1}} + \frac{1}{k _{2}}$
$\Rightarrow k_{e q} = \frac{k _{1} k _{2}}{k _{1} + k _{2}}$
At equilibrium,
$m g = k_{e q} x$
$\Rightarrow x = \frac{m g ( k _{1} + k _{2} )}{k _{1} k _{2}}$

Q. A mass M is suspended by two springs A and B of force constants k1​ and k2​ respectively as shown in the diagram. The total stretch of springs in equilibrium is:

k1​+k2​Mg​

(k1​+k2​)2Mg​

2(k1​+k2​)Mg​

k1​k2​Mg(k1​+k2​)​