Spring-Mass System: Analyzing Natural Frequencies

This problem involves calculating the natural frequencies of a spring-mass system under two different configurations: springs in series and springs in parallel. We will then determine the ratio of these natural frequencies. A clear understanding of equivalent stiffness for different spring arrangements is crucial here.

Spring Stiffness Unit Conversion

First, let's convert the given spring stiffness values from N/mm to N/m, which is the standard SI unit for stiffness, as mass is given in kg.

• Stiffness of first spring, $$\small k_1 = 8 \text{ N/mm}$$
• To convert N/mm to N/m, we multiply by 1000 (since 1 m = 1000 mm):
• $$\small k_1 = 8 \times 1000 \text{ N/m} = 8000 \text{ N/m}$$
• Stiffness of second spring, $$\small k_2 = 12 \text{ N/mm}$$
• $$\small k_2 = 12 \times 1000 \text{ N/m} = 12000 \text{ N/m}$$

The mass of the system is given as $$\small m = 5 \text{ kg}$$.

Natural Frequency for Springs in Series (Case i)

In case (i), the mass is suspended at the bottom of two springs connected in series.

Equivalent Stiffness Calculation for Series Connection

For springs connected in series, the reciprocal of the equivalent stiffness ($$\small k_{eq,i}$$) is the sum of the reciprocals of individual stiffnesses.

The formula for equivalent stiffness in series is:
$$\small \frac{1}{k_{eq,i}} = \frac{1}{k_1} + \frac{1}{k_2}$$

Substituting the values:
$$\small \frac{1}{k_{eq,i}} = \frac{1}{8000 \text{ N/m}} + \frac{1}{12000 \text{ N/m}}$$ $$\small \frac{1}{k_{eq,i}} = \frac{3}{24000} + \frac{2}{24000}$$ $$\small \frac{1}{k_{eq,i}} = \frac{5}{24000}$$ $$\small k_{eq,i} = \frac{24000}{5} = 4800 \text{ N/m}$$

Natural Frequency Calculation for Case (i)

The natural frequency ($$\small \omega_n$$) of a spring-mass system is given by the formula:
$$\small \omega_n = \sqrt{\frac{k_{eq}}{m}}$$

For case (i), the natural frequency ($$\small \omega_{n,i}$$) is:
$$\small \omega_{n,i} = \sqrt{\frac{k_{eq,i}}{m}} = \sqrt{\frac{4800 \text{ N/m}}{5 \text{ kg}}}$$ $$\small \omega_{n,i} = \sqrt{960} \text{ rad/s}$$ $$\small \omega_{n,i} \approx 30.9838 \text{ rad/s}$$

Natural Frequency for Springs in Parallel (Case ii)

In case (ii), the mass is fixed between two springs. This arrangement typically implies a parallel connection, where both springs contribute directly to the restoring force on the mass.

Equivalent Stiffness Calculation for Parallel Connection

For springs connected in parallel, the equivalent stiffness ($$\small k_{eq,ii}$$) is the sum of the individual stiffnesses.

The formula for equivalent stiffness in parallel is:
$$\small k_{eq,ii} = k_1 + k_2$$

Substituting the values:
$$\small k_{eq,ii} = 8000 \text{ N/m} + 12000 \text{ N/m}$$ $$\small k_{eq,ii} = 20000 \text{ N/m}$$

Natural Frequency Calculation for Case (ii)

For case (ii), the natural frequency ($$\small \omega_{n,ii}$$) is:
$$\small \omega_{n,ii} = \sqrt{\frac{k_{eq,ii}}{m}} = \sqrt{\frac{20000 \text{ N/m}}{5 \text{ kg}}}$$ $$\small \omega_{n,ii} = \sqrt{4000} \text{ rad/s}$$ $$\small \omega_{n,ii} \approx 63.2456 \text{ rad/s}$$

Ratio of Natural Frequencies

We need to find the ratio of natural frequencies of case (ii) to those of case (i).

Ratio $$= \frac{\omega_{n,ii}}{\omega_{n,i}}$$ Ratio $$= \frac{\sqrt{4000}}{\sqrt{960}}$$ Ratio $$= \sqrt{\frac{4000}{960}}$$ Ratio $$= \sqrt{\frac{400}{96}}$$ Ratio $$= \sqrt{\frac{100}{24}}$$ Ratio $$= \sqrt{\frac{25}{6}}$$ Ratio $$\approx \sqrt{4.16666...}$$ Ratio $$\approx 2.0412$$

Conclusion and Final Answer

The calculated ratio of natural frequencies of case (ii) to case (i) is approximately 2.0412.

Comparing this value with the given options:

The closest approximation to 2.0412 is 2.0.

Therefore, the ratio of natural frequencies of the case (ii) to those of case (i) is approximately 2.0.