Question

Two identical springs, each of spring factor k, may be connected in the following ways. Deduce the spring factor of the oscillation of the body in each case.

Answer 1

For each spring

F = -ky .........(i)

where, F = restoring force, k = spring factor and y = displacement of the spring

(i) In Fig. (a), let the mass m produce a displacement y in each spring and F be the restoring force in each spring, if k₁ be the spring factor of the combined systems, then

2F = -k₁y

or F \( =-\frac{k_1}{2}y\ ...(ii)\)   

Comparing (i) and (ii), we get

\(\frac{k_1}{2} = k\ or\ k_1 = 2k\)

(ii) If Fig. (b), as the length of the spring is doubled, the mass m will produce double the displacement (2y). If k₂ be the spring factor of the combined system then

F = -k₂ (2y) = -2 k₂y .........(iii)

Comparing (i) and (iii), we get

2k₂ = k or k₂ = \(\frac{k}{2}\) 

(iii) In Fig. (c), the mass m stretches the upper spring and compresses the lower spring, each giving rise to a restoring force F in the same direction. If k₃ be the spring factor of the combined system, then

2F = -k₃ y

or F = \(-\frac{k_3}{2}\) y ........(iv)

Comparing (i) and (iv)

\(\frac{k_3}{2}\) = k or k₃ = 2k