Question

When a mass m is attached to a spring it exerts a force W = mg on the spring and the length of the spring is changed by delta x. If the single spring is replaced with a) two identical springs in series, what happens to delta xseries compared to the case of a single spring? b) If the single spring is replaced by two identical springs in parallel, what happens to delta xparallel compared to the case of a single spring?

See figure above. Assume all springs are identical, i.e. have the same spring constant k, length, mass, etc. Answer questions a) and b) by stating if delta x increases, decreases or remains unchanged and compare it to the single spring case, i.e. what are delta xseries and delta xparallel in terms of delta x for the single spring case? Hint: Draw a force diagram of the system remembering that the net force on the mass must be zero when it is in equilibrium.

The figure is second from last page of this website: http://www.pa.msu.edu/courses/2015summer/PHY251/labfiles/exp9.pdf

Answer 1

When a mass is attached to a spring it exerts a force on the spring and the length of the spring is changed by x.

(a) For a single spring :

see a free body diagram in figure 1 below,

by Hooke's law,                F = - k x                       { eq.1 }

by newton's second law,     F = - m g                                      { eq.2 }

where, g = acceleration due to gravity in vertical direction (-ve)

comparing above two equations,

k x = m g

or slope, k = g m / x                      { eq. 3 }

where, x = length of the spring

from a slope, To obtain a graph between m vs. x.

(b) Spring connected in series :

from a free body diagram which can be shown in figure 2 below,

for spring 1, k₁ x₁ = mg

or x₁ = mg / k₁

similarly as : for spring 2,   k₂ x₂ = mg

or x₂ = mg / k₂

total distance for the series combination is given as :

x = x₁ + x₂                      { eq. 4 }

and the effective spring constant in series is given as :

k_s x = mg                            { eq. 5 }

inserting the value of 'x' in eq.5,

k_s ( x₁ + x₂ ) = mg

k_s [ (mg / k₁) + (mg / k₂) ] = mg

1 / k_s = 1 / k₁ + 1 / k₂

(c) spring connected in parallel :

see a free body diagram which can be shown below,

spring which connected in parallel given as ::

k₁ x₁ + k₂ x₂ = mg = k_p x                      { eq.6 }

where, k_p = effective spring constant in parallel

k_p = (k₁ + k₂)