In the reported figure, two bodies A and B of masses 200 g and 800 g are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be ___ rad s$$^{-1}$$ when $$k = 20$$ N m$$^{-1}$$.

We need to determine the angular frequency ($$\omega$$) of the two-body spring system when it is released from its stretched position on a frictionless horizontal surface.

1. Analyze the Spring Combination

From the problem , two springs ($$S_1$$ and $$S_2$$) are connected end-to-end between block $$A$$ and block $$B$$. This configuration constitutes a series combination.

• Stiffness of first spring ($$k_1$$) = $$k$$
• Stiffness of second spring ($$k_2$$) = $$4k$$

We calculate the equivalent spring constant ($$k_{\text{eq}}$$) using the series formula:

$$\frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{1}{k} + \frac{1}{4k}$$

$$\frac{1}{k_{\text{eq}}} = \frac{4 + 1}{4k} = \frac{5}{4k} \implies k_{\text{eq}} = \frac{4k}{5}$$

Given that $$k = 20\text{ N m}^{-1}$$, we find the numerical value of $$k_{\text{eq}}$$:

$$k_{\text{eq}} = \frac{4 \times 20}{5} = 16\text{ N m}^{-1}$$

2. Calculate the Reduced Mass ($$\mu$$)

Since both masses are free to move on the frictionless surface, the system acts as a two-body oscillator. The motion can be simplified by analyzing it using the concept of reduced mass ($$\mu$$):

• Mass of body $$A$$ ($$m_A$$) = $$200\text{ g} = 0.2\text{ kg}$$
• Mass of body $$B$$ ($$m_B$$) = $$800\text{ g} = 0.8\text{ kg}$$

$$\mu = \frac{m_A \cdot m_B}{m_A + m_B} = \frac{0.2 \times 0.8}{0.2 + 0.8} = \frac{0.16}{1.0} = 0.16\text{ kg}$$

3. Determine the Angular Frequency ($$\omega$$)

The formula for the angular frequency of a two-body spring system is given by:

$$\omega = \sqrt{\frac{k_{\text{eq}}}{\mu}}$$

Substitute our calculated values for $$k_{\text{eq}}$$ and $$\mu$$ into the equation:

$$\omega = \sqrt{\frac{16}{0.16}} = \sqrt{100} = 10\text{ rad s}^{-1}$$

Conclusion

The angular frequency of the system when released is 10 $$\text{rad s}^{-1}$$.