# 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 (

Mass (m) = 160 gram = 0.16 kg

Acceleration due to gravity (g) = 10 m/s^{2}

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 (k_{2}) and spring 3 (k_{3}) tare connected in parallel. The equivalent constant :

k_{23} = k_{2} + k_{3} = 40 + 40 = 80 N/m

Spring 1 (k_{1}) and spring 23 (k_{23}) are connected in series. The equivalent constant :

1/k = 1/k_{1} + 1/k_{23} = 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

_{1 }= k

_{2}= k

_{3 }= 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 k_{1 }= k_{2} = k_{3 }= 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 (k_{1}) and spring 2 (k_{2}) are connected in parallel. The equivalent constant :

k_{12} = k_{1} + k_{2} = 300 + 300 = 600 N/m

Spring 3 (k_{3}) and spring 12 (k_{12}) are connected in series. The equivalent constant :

1/k = 1/k_{3} + 1/k_{12} = 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 (k_{1}) = k = 50 Nm^{-1}

Spring constant 2 (k_{2}) = k = 50 Nm^{-1}

Spring constant 3 (k_{3}) = 2k = 2 (50 Nm^{-1}) = 100 Nm^{-1}

Object’s mass (m) = 400 gram = 0.4 kg

Acceleration due to gravity (g) = 10 m/s^{2}

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

__Wanted :__ The change in length (Δx)

__Solution :__

Determine the equivalent constant :

Spring 1 (k_{1}) and spring 2 (k_{2}) are connected in parallel. The equivalent constant :

k_{12} = k_{1} + k_{2} = 50 + 50 = 100 N/m

Spring 3 (k_{3}) and spring 12 (k_{12}) are connected in series. The equivalent constant :

1/k = 1/k_{3} + 1/k_{12} = 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