What is the effective spring constant if two springs of spring constant k1 and k2 are connected in series and parallel?

You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.

  • Thread starter shepherd882
  • Start date Sep 27, 2018

a) Consider 2 springs, connected in series. If they have different spring constants k1 and k2 then what is the effective spring constant for the double spring system? Give a convincing argument for your formula. You may assume that the mass of the springs is negligible. b) Now suppose that 2 springs are hanging in parallel. (Assume that they are connected to the same point on a stand and on a hanging weight, so that they both stretch by the same amount.) They both have the same unstretched length but different spring constants k1 and k2. What is the effective spring constant for this double spring system? Again, give a convincing argument. Fig 4.15: c) Now suppose that 3 springs are connected in series, with spring constants k1, k2 and k3. What is the effective spring constant in this case? d) Compare your formula for springs in series and parallel to the formulas for electrical resistances in series and parallel. THANK YOU

Homework Equations

F = ks ka <-- effective spring constant k <-- spring constant

The Attempt at a Solution

a) equation 1 - F=k1s1 = k2s2 equation 2 - F = ka( s1 + s2) combined equation: ka = (k1k2)/(k1+k2)

not sure what to do for part b and c and kinda have an idea for part d

Answers and Replies

CWatters

Hints: b) each spring produces a different force (if the spring constants are different). Do you know how to add two forces to make one equivalent force? c) extend your answer for a)

d) How do you add resistors in parallel? Compare with your equation for springs in series.

for part b) would it be like this: equation 1: Fnet = k1s + k2s equation 2: F = kas combined equation: ka = k1 + k2 Thank you for your help!!

Hints: b) each spring produces a different force (if the spring constants are different). Do you know how to add two forces to make one equivalent force? c) extend your answer for a)

d) How do you add resistors in parallel? Compare with your equation for springs in series.

Parallel.

When two massless springs following Hooke's Law, are connected via a thin, vertical rod as shown in the figure below, these are said to be connected in parallel. Spring 1 and 2 have spring constants #k_1# and #k_2# respectively. A constant force #vecF# is exerted on the rod so that remains perpendicular to the direction of the force. So that the springs are extended by the same amount. Alternatively, the direction of force could be reversed so that the springs are compressed.

What is the effective spring constant if two springs of spring constant k1 and k2 are connected in series and parallel?

This system of two parallel springs is equivalent to a single Hookean spring, of spring constant #k#. The value of #k# can be found from the formula that applies to capacitors connected in parallel in an electrical circuit.

#k=k_1+k_2#

Series.

When same springs are connected as shown in the figure below, these are said to be connected in series. A constant force #vecF# is applied on spring 2. So that the springs are extended and the total extension of the combination is the sum of elongation of each spring. Alternatively, the direction of force could be reversed so that the springs are compressed.

What is the effective spring constant if two springs of spring constant k1 and k2 are connected in series and parallel?

This system of two springs in series is equivalent to a single spring, of spring constant #k#. The value of #k# can be found from the formula that applies to capacitors connected in series in an electrical circuit.

For spring 1, from Hooke's Law

#F=k_1x_1#

where #x_1# is the deformation of spring.

Similarly if #x_2# is the deformation of spring 2 we have

#F=k_2x_2#

Total deformation of the system

#x_1+x_2=F/k_1+F/k_2#
#=>x_1+x_2=F(1/k_1+1/k_2)#

Rewriting and comparing with Hooke's law we get

#k=(1/k_1+1/k_2)^-1#