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Question: What is the difference between the series and parallel combination of springs? Calculate the value o...

What is the difference between the series and parallel combination of springs? Calculate the value of effective spring constant for every combination.

Explanation

Solution

When a spring is stretched or compressed, it undergoes a displacement x which is directly proportional to the restoring force F. This is mathematically expressed as FxF \propto x . Removing the proportionality sign we get, F=kxF = - kx .Where k is the spring constant of the spring which gives the measure of the stiffness of the spring. The negative sign denotes that the nature of force is restoring.

Complete step-by-step answer:
The spring constant of the spring which gives the measure of the stiffness of the spring. The restoring force is given by F=kxF = - kx .
There are two possible combinations in which the springs can be connected.
Series combination: Here the springs are attached end to end in a chain fashion
Parallel combination: Here the springs originate and terminate at the same points
In the series combination,
Say there are two springs S1,S2{S_{1\,}},\,{S_2} connected in a series combination.
The restoring force on spring S1{S_{1\,}} will be F1=k1x1{F_1} = - {k_1}{x_1} and the restoring force on spring S2{S_2} will be F2=k2x2{F_2} = - {k_2}{x_2} .
Since the tension acting in both the springs is the same, F1=F2=F{F_1} = {F_2} = F where F is the equivalent force.
Now the displacements are given as x1=F1k1{x_1} = \dfrac{{{F_1}}}{{{k_1}}} and x2=F2k2{x_2} = \dfrac{{{F_2}}}{{{k_2}}} . the equivalent displacement is given as
x=Fkx = \dfrac{F}{k} .
The equivalent displacement is the sum total of all the individual displacements.
So, x=x1+x2x = {x_1} + {x_2} .
Substituting the values,
x=F1x1+F2x2x = \dfrac{{{F_1}}}{{{x_1}}} + \dfrac{{{F_2}}}{{{x_2}}}
Fx=F1x1+F2x2\Rightarrow \dfrac{F}{x} = \dfrac{{{F_1}}}{{{x_1}}} + \dfrac{{{F_2}}}{{{x_2}}}
We know that
F1=F2=F{F_1} = {F_2} = F
Hence the equation reduces to Fx=Fx1+Fx2\dfrac{F}{x} = \dfrac{F}{{{x_1}}} + \dfrac{F}{{{x_2}}}
Further solving this we get,
1x=1x1+1x2\dfrac{1}{x} = \dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}}
This is the formula for the springs connected in series.
In the parallel combination,
Say there are two springs S1,S2{S_{1\,}},\,{S_2} connected in a parallel combination.
The restoring force on spring S1{S_{1\,}} will be F1=k1x1{F_1} = - {k_1}{x_1} and the restoring force on spring S2{S_2} will be F2=k2x2{F_2} = - {k_2}{x_2} .
Since the sum of tension acting in both the springs is the equivalent force, F1+F2=F{F_1} + {F_2} = F where F is the equivalent force.
Substituting the values,
k1x1+k2x2=kx{k_1}{x_1} + {k_2}{x_2} = kx
Since the displacement is the same for each spring, x1=x2=x{x_1} = {x_2} = x .
The equation reduces to k1x+k2x=kx{k_1}x + {k_2}x = kx
k1+k2=k\Rightarrow {k_1} + {k_2} = k
This is the formula for the springs connected in parallel.

Note: The formula for the series and parallel combinations in the springs is just the opposite as in the case of resistances. So, the difference must be noted carefully. The negative sign only denotes the nature of the force. Thus, it can be omitted in calculations.