Question

The ratio of spring constant of two springs is 2:3 .What is the ratio of potential energy of springs if the are stretched by same force??

Answer 1

Given :

The ratio of spring constant of two springs is 2:3 and the springs are stretched by same force.

To Find :

The ratio of potential energy of springs

Solution :

 F = kx (where k = spring constant, and x = Displacement of the spring from its equilibrium position)

Let the spring constants for two springs be 2a and 3a respectively.

For spring 1,

F = $K_{1} x_{1}$

or, $x_{1}$ = $\frac{F}{K_{1} }$

∴   $x_{1}$ = $\frac{F}{2a}$

For spring 2,

F = $K_{2} x_{2}$

or, $x_{2}$ = $\frac{F}{K_{2} }$

∴   $x_{2}$ = $\frac{F}{3a}$

We know, Potential Energy in a stretched spring = $\frac{1}{2} Kx^{2}$

Potential Energy stored in spring 1 (U$_{1}$) = $\frac{1}{2} K_{1} x_{1} ^{2}$

                                                          U$_{1}$   =  $\frac{1}{2} *2a * (\frac{F}{2a}) ^{2}$

                                                          U$_{1}$   = $\frac{F^{2}a }{4a^{2} }$

                                                          U$_{1}$   = $\frac{F^{2} }{4a}$

Potential Energy stored in spring 2 (U$_{2}$) = $\frac{1}{2} K_{2} x_{2} ^{2} }$

                                                           U$_{2}$  = $\frac{1}{2} *3a * (\frac{F}{3a}) ^{2}$

                                                           U$_{2}$  = $\frac{F^{2} }{6a}$

∴    U$_{1}$  :  U$_{2}$

or,  $\frac{F^{2} }{4a}$  :  $\frac{F^{2} }{6a}$

∴    3   :   2

∴  The ratio of potential energy of springs when they are stretched by same force is 3:2.

Answer 2

Given:

Ratio of spring constants of two springs $= 2:3$

To find:

Ratio of Potential energy of the springs.

Solution:

Step 1

According to the question, two springs with spring constant $k_{1}$ and $k_{2}$ act under forces $F_{1}$ and $F_{2}$ to produce extensions $x_{1}$ and $x_{2}$ in the springs.

According to Stokes Law, for a force on spring, we have

$F=-kx$

Where, k is the spring constant.

For spring 1, we have

$F_{1}= k_{1} x_{1}$      $(i)$

Similarly for spring 2

$F_{2}=k_{2} x_{2}$     $(ii)$

But, we have been given that the springs act under the same force

Hence, $F_{1} =F_{2}$  , therefore, equating both the values of forces we get

$k_{1} x_{1}= k_{2} x_{2}$

$\frac{k_{1} }{k_{2} }= \frac{x_{2} }{x_{1} }$

We know, $\frac{k_{1} }{k_{2} }= \frac{2}{3}$

Therefore, $\frac{x_{1} }{x_{2} }= \frac{3}{2}$  $(iii)$

Step 2

Now,

The Potential energy (P.E.) of a stretched spring is given by $U=\frac{1}{2}kx^{2}$  

Hence,

P.E. of spring 1 is  $U_{1} =\frac{1}{2}k_{1}x_{1} ^{2}$    $;$ $and$

P.E. of spring 2 is $U_{2} =\frac{1}{2}k_{2}x_{2} ^{2}$

For the ratio of Potential energies of the 2 springs,

$\frac{U_{1} }{U_{2} }= \frac{\frac{1}{2} k_{1} x_{1} ^{2} }{\frac{1}{2}k_{2}x_{2}^{2} }$   $;$ $or$

$\frac{U_{1} }{U_{2} }= \frac{k_{1} x_{1}^{2} }{k_{2} x_{2}^{2} }$  

We know, $\frac{k_{1} }{k_{2} }= \frac{2}{3}$  $and$  $\frac{x_{1} }{x_{2} }= \frac{3}{2}$

Substituting this value in the equation, we get

$\frac{U_{1} }{U_{2} }=\frac{k_{1} }{k_{2} }(\frac{x_{1} }{x_{2} } )^{2}$

Hence,

$\frac{U_{1} }{U_{2} }=\frac{2}{3}(\frac{3}{2} )^{2}$

Therefore,

$\frac{U_{1} }{U_{2} }=\frac{3}{2}$

Final answer:

Hence, the ratio of potential energy of the springs will be $3:2$.