Springs in series and parallel – problems and solutions

1. A 160-gram object attaches at one end of a spring and the change in length of the spring is 4 cm. What is the change in length of three springs connected in series and parallel, as shown in the figure below?

Known :

The change in length of a spring (Δx) = 4 cm = 0.04 m Springs in series and parallel – problems and solutions 1

Mass (m) = 160 gram = 0.16 kg

Acceleration due to gravity (g) = 10 m/s²

Weight (w) = m g = (0.16)(10) = 1.6 Newton

Wanted : The change in length of three spring (Δx)

Solution :

The equation of Hooke’s law :

k = w / Δx = 1.6 / 0.04 = 40 N/m

The three springs have the same constant, k = 40 N/m.

Determine the equivalent constant :

Spring 2 (k₂) and spring 3 (k₃) tare connected in parallel. The equivalent constant :

k₂₃ = k₂ + k₃ = 40 + 40 = 80 N/m

Spring 1 (k₁) and spring 23 (k₂₃) are connected in series. The equivalent constant :

1/k = 1/k₁ + 1/k₂₃ = 1/40 + 1/80 = 2/80 + 1/80 = 3/80

k = 80/3

Determine the change in length of three springs :

Δx = w / k = 1.6 : 80/3 = (1.6)(3/80) = 4.8 / 80 = 0.06 m = 6 cm

2. Three springs with the same constant connected in series and parallel, and a 2-kg object attached at one end of a spring, as shown in figure below. Spring constant is k₁= k₂ = k₃= 300 N/m. What is the change in length of the three springs. Acceleration due to gravity is g = 10 m.s^-2.

Known :

Spring constant k₁= k₂ = k₃= 300 N.m^-1

Acceleration due to gravity (g) = 10 m.s^-2

Object’s mass (m) = 2 kg

Object’s weight (w) = m g = (2)(10) = 20 Newton

Wanted : The change in length of the three springs (Δx)

Solution :

Determine the equivalent constant :

Spring 1 (k₁) and spring 2 (k₂) are connected in parallel. The equivalent constant :

k₁₂ = k₁ + k₂ = 300 + 300 = 600 N/m

Spring 3 (k₃) and spring 12 (k₁₂) are connected in series. The equivalent constant :

1/k = 1/k₃ + 1/k₁₂ = 1/300 + 1/600 = 2/600 + 1/600 = 3/600

k = 600/3 = 200 N/m

Determine the change in length of the three springs :

Δx = w / k = 20/200 = 2/20 = 1/10 = 0.1 m

3. Three springs are connected in series and parallel, as shown in figure below. If spring constant k = 50 Nm^-1 and a mass of 400 gram attached at one end of a spring. What is the change in length of the three springs.

Known :

Spring constant 1 (k₁) = k = 50 Nm^-1

Spring constant 2 (k₂) = k = 50 Nm^-1

Spring constant 3 (k₃) = 2k = 2 (50 Nm^-1) = 100 Nm^-1

Object’s mass (m) = 400 gram = 0.4 kg

Acceleration due to gravity (g) = 10 m/s²

Object’s weight (w) = m g = (0.4)(10) = 4 Newton

Wanted : The change in length (Δx)

Solution :

Determine the equivalent constant :

Spring 1 (k₁) and spring 2 (k₂) are connected in parallel. The equivalent constant :

k₁₂ = k₁ + k₂ = 50 + 50 = 100 N/m

Spring 3 (k₃) and spring 12 (k₁₂) are connected in series. The equivalent constant :

1/k = 1/k₃ + 1/k₁₂ = 1/100 + 1/100 = 2/100

k = 100/2 = 50 N/m

Determine the change in length of the three springs :

Δx = w / k = 4 / 50 = = 0.08 m = 8 cm

How does combining springs in series affect the overall spring constant?
- Answer: When springs are combined in series, the overall spring constant is reduced. The reciprocal of the equivalent spring constant is the sum of the reciprocals of the individual spring constants: 1/k $=1/ k ^{1} + 1/ k ^{2}$ .
How does the overall spring constant change when springs are combined in parallel?
- Answer: Combining springs in parallel results in an overall spring constant that is the sum of the individual spring constants: k $= k_{1} + k_{2}$ .
If two identical springs are arranged in series, how does the combined spring constant compare to the spring constant of an individual spring?
- Answer: The combined spring constant will be half of the spring constant of one of the individual springs.

What happens to the extension or compression of springs in series when a force is applied?
- Answer: For springs in series, the same force causes each spring to extend or compress, but the total extension (or compression) is the sum of the extensions (or compressions) of the individual springs.
If springs in parallel are subjected to a force, how is that force distributed?
- Answer: For springs in parallel, the force is distributed among the springs based on their spring constants. Springs with a higher spring constant will bear a greater portion of the force than those with a lower spring constant.
Why can springs in series be thought of as a single spring with a longer length?
- Answer: Springs in series have a combined effect equivalent to stretching a single longer spring. The extensions or compressions of the individual springs add up, just as they would in a longer singular spring.

How does the potential energy stored in springs in series compare to that in springs in parallel for the same applied force?
- Answer: Springs in series store more potential energy than springs in parallel for the same applied force because they undergo a greater combined extension or compression.
If one of the springs in a parallel configuration breaks or becomes ineffective, what happens to the overall behavior of the system?
- Answer: If one spring in a parallel configuration breaks, the remaining springs will still function. However, the overall spring constant of the system will decrease, and the system won’t be able to exert as much restoring force as before.
Why are springs in series more susceptible to larger deformations than those in parallel for the same applied force?
- Answer: For springs in series, the same force acts on each spring, causing each one to extend or compress. The total deformation is the sum of the individual deformations. In parallel, the force is distributed among the springs, so each one experiences a reduced effective force, leading to smaller individual deformations.

In practical applications, why might engineers choose to use springs in parallel rather than in series?

Answer: Engineers might choose springs in parallel to achieve a higher overall spring constant, resulting in stiffer behavior. This setup can also provide redundancy; if one spring fails, the system continues to function, albeit with a reduced overall spring constant.

Related posts: