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What is the difference between the series and parallel combination of springs? Calculate the value of effective spring constant for every combination.
Answer
397.2k+ views
Hint: 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 $F \propto x$ . Removing the proportionality sign we get, $F = - kx$ .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 $F = - kx$ .
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 ${S_{1\,}},\,{S_2}$ connected in a series combination.
The restoring force on spring ${S_{1\,}}$ will be ${F_1} = - {k_1}{x_1}$ and the restoring force on spring ${S_2}$ will be ${F_2} = - {k_2}{x_2}$ .
Since the tension acting in both the springs is the same, ${F_1} = {F_2} = F$ where F is the equivalent force.
Now the displacements are given as ${x_1} = \dfrac{{{F_1}}}{{{k_1}}}$ and ${x_2} = \dfrac{{{F_2}}}{{{k_2}}}$ . the equivalent displacement is given as
$x = \dfrac{F}{k}$ .
The equivalent displacement is the sum total of all the individual displacements.
So, $x = {x_1} + {x_2}$ .
Substituting the values,
$x = \dfrac{{{F_1}}}{{{x_1}}} + \dfrac{{{F_2}}}{{{x_2}}}$
$ \Rightarrow \dfrac{F}{x} = \dfrac{{{F_1}}}{{{x_1}}} + \dfrac{{{F_2}}}{{{x_2}}}$
We know that
${F_1} = {F_2} = F$
Hence the equation reduces to $\dfrac{F}{x} = \dfrac{F}{{{x_1}}} + \dfrac{F}{{{x_2}}}$
Further solving this we get,
$\dfrac{1}{x} = \dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}}$
This is the formula for the springs connected in series.
In the parallel combination,
Say there are two springs ${S_{1\,}},\,{S_2}$ connected in a parallel combination.
The restoring force on spring ${S_{1\,}}$ will be ${F_1} = - {k_1}{x_1}$ and the restoring force on spring ${S_2}$ will be ${F_2} = - {k_2}{x_2}$ .
Since the sum of tension acting in both the springs is the equivalent force, ${F_1} + {F_2} = F$ where F is the equivalent force.
Substituting the values,
${k_1}{x_1} + {k_2}{x_2} = kx$
Since the displacement is the same for each spring, ${x_1} = {x_2} = x$ .
The equation reduces to ${k_1}x + {k_2}x = kx$
$ \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.
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 $F = - kx$ .
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 ${S_{1\,}},\,{S_2}$ connected in a series combination.
The restoring force on spring ${S_{1\,}}$ will be ${F_1} = - {k_1}{x_1}$ and the restoring force on spring ${S_2}$ will be ${F_2} = - {k_2}{x_2}$ .
Since the tension acting in both the springs is the same, ${F_1} = {F_2} = F$ where F is the equivalent force.
Now the displacements are given as ${x_1} = \dfrac{{{F_1}}}{{{k_1}}}$ and ${x_2} = \dfrac{{{F_2}}}{{{k_2}}}$ . the equivalent displacement is given as
$x = \dfrac{F}{k}$ .
The equivalent displacement is the sum total of all the individual displacements.
So, $x = {x_1} + {x_2}$ .
Substituting the values,
$x = \dfrac{{{F_1}}}{{{x_1}}} + \dfrac{{{F_2}}}{{{x_2}}}$
$ \Rightarrow \dfrac{F}{x} = \dfrac{{{F_1}}}{{{x_1}}} + \dfrac{{{F_2}}}{{{x_2}}}$
We know that
${F_1} = {F_2} = F$
Hence the equation reduces to $\dfrac{F}{x} = \dfrac{F}{{{x_1}}} + \dfrac{F}{{{x_2}}}$
Further solving this we get,
$\dfrac{1}{x} = \dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}}$
This is the formula for the springs connected in series.
In the parallel combination,
Say there are two springs ${S_{1\,}},\,{S_2}$ connected in a parallel combination.
The restoring force on spring ${S_{1\,}}$ will be ${F_1} = - {k_1}{x_1}$ and the restoring force on spring ${S_2}$ will be ${F_2} = - {k_2}{x_2}$ .
Since the sum of tension acting in both the springs is the equivalent force, ${F_1} + {F_2} = F$ where F is the equivalent force.
Substituting the values,
${k_1}{x_1} + {k_2}{x_2} = kx$
Since the displacement is the same for each spring, ${x_1} = {x_2} = x$ .
The equation reduces to ${k_1}x + {k_2}x = kx$
$ \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.
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