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 mSprings 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

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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-1Springs in series and parallel – problems and solutions 2

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

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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-1Springs in series and parallel – problems and solutions 3

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

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