A spring-mass system consists of a mass of 5 kg and two springs of stiffness 8 N/mm and 12 N/mm. The system is arranged in different manners, that is:

(i) the mass is suspended at the bottom of two springs in series, and

(ii) the mass is fixed between two springs.

The ratio of natural frequencies of the case (ii) to those of case (i) is approximately ___________. 

Concept:-

Springs in series - 

Consider two springs with force constants k₁ and k₂ connected in series supporting a load,

F = mg. 

The equivalent stiffness of the system can be written as,

⇒ \(\frac{1}{k_s}=\frac{1}{k_1}+\frac{1}{k_2}\;\)

Springs in Parallel -

Consider two springs with force constants k₁ and k₂ connected in parallel, supporting a load F = mg. 

The equivalent stiffness of the system can be written as,

⇒ k_p = k₁ + k₂ 

Natural frequency for an undamped spring-mass system can be written as,

⇒ \(\omega_n = \sqrt{\frac{k}{m}}\;\)

Calculation:-

Given:-

m = 5 kg, k1 = 8 N/mm, k2 = 12 N/mm

Case(1) - The mass is suspended at the bottom of two springs in series -

Equivalent stiffness is, \(\frac{1}{k_s}=\frac{1}{k_1}+\frac{1}{k_2}\;\)

So, \(\frac{1}{k_s}=\frac{1}{8}+\frac{1}{12}\;\)

k_s = 4.8 N/mm = 4800 N/m

Then natural frequency is, \(\omega_n = \sqrt{\frac{k_s}{m}}\;\) 

\(\omega_{n1} = \sqrt{\frac{4.8×10^3}{5}}= 30.983\;rad/s\;\)

Case(2) - The mass is fixed between two springs.

Equivalent stiffness is, k_p = k₁ + k₂

So, k_p = 8 + 12 = 20 N/mm = 20 × 10³ N/m

Then natural frequency is, \(\omega_n = \sqrt{\frac{k_p}{m}}\;\)

⇒ \(\omega_{n2}= \sqrt{\frac{20000}{5}}= 63.245\;rad/s\;\)

The ratio of natural frequencies of case (ii) to those of case (i) is,

⇒ \(\frac{\omega _{n2}}{\omega _{n1}}= \frac{63.245}{30.983}=2.04\;\)≈ 2