When a tensile or compressive load `'P'` is applied to rod or cable, its length changes. The change length `x` which, for an elastic material is proportional to the force Hook' law).
`P alpha x` or `P = kx`
The above equation is similar to the equation of spring. For a rod of length `L`, area `A` and young modulue `Y`. the extension `x`can be expressed as.
`x = (PL)/(AY)` or `P = (AY)/(x)`, hence `K = (AY)/(L)`
Thus rod or cables attached to lift can be treated as springs. The energy stored in rod is called stra energy & equal to `(1)/(2)Px`. The loads placed or dropped on the floor of lift cause stresses in the cable and can be evaluted by spring analogy. If the cable of lift cause stresses and load is placed dropped, then maximum extension in cable can be calculated by enerfy conservation.
It two rods of same length `(4m)` and cross section areas `2 cm^(2)` and `4 cm^(2)` with same young modulus `2 xx 10^(10)N//m^(2)` are attached one after the other with mass `600 kg` then angular frequency is-

Options

A. `(1000)/(3)`
B. `(10)/(3)`
C. `(100)/(3)`
D. `(10pi)/(3)`

Explanation

To solve the problem step by step, we will follow the concepts of elasticity and the analogy of springs.

### Step 1: Determine the spring constants for both rods.
We know that the spring constant \( K \) for a rod can be expressed as:
\[ K = \frac{AY}{L} \]
where:
- \( A \) = cross-sectional area
- \( Y \) = Young's modulus
- \( L \) = length of the rod

#### For Rod 1:
- Length \( L_1 = 4 \, \text{m} \)
- Area \( A_1 = 2 \, \text{cm}^2 = 2 \times 10^{-4} \, \text{m}^2 \)
- Young's modulus \( Y = 2 \times 10^{10} \, \text{N/m}^2 \)

Calculating \( K_1 \):
\[ K_1 = \frac{(2 \times 10^{-4} \, \text{m}^2)(2 \times 10^{10} \, \text{N/m}^2)}{4 \, \text{m}} = \frac{4 \times 10^{6} \, \text{N}}{4} = 10^6 \, \text{N/m} \]

#### For Rod 2:
- Length \( L_2 = 4 \, \text{m} \)
- Area \( A_2 = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \)

Calculating \( K_2 \):
\[ K_2 = \frac{(4 \times 10^{-4} \, \text{m}^2)(2 \times 10^{10} \, \text{N/m}^2)}{4 \, \text{m}} = \frac{8 \times 10^{6} \, \text{N}}{4} = 2 \times 10^{6} \, \text{N/m} \]

### Step 2: Find the effective spring constant for the two rods in series.
When two springs (or rods in this case) are connected in series, the effective spring constant \( K_{\text{effective}} \) is given by:
\[ \frac{1}{K_{\text{effective}}} = \frac{1}{K_1} + \frac{1}{K_2} \]

Substituting the values:
\[ \frac{1}{K_{\text{effective}}} = \frac{1}{10^6} + \frac{1}{2 \times 10^6} = \frac{2}{2 \times 10^6} + \frac{1}{2 \times 10^6} = \frac{3}{2 \times 10^6} \]
Thus,
\[ K_{\text{effective}} = \frac{2 \times 10^6}{3} \, \text{N/m} \]

### Step 3: Calculate the angular frequency \( \omega \).
The angular frequency \( \omega \) for a mass-spring system is given by:
\[ \omega = \sqrt{\frac{K_{\text{effective}}}{m}} \]
where \( m \) is the mass attached to the rods.

Given:
- Mass \( m = 600 \, \text{kg} \)

Substituting the values:
\[ \omega = \sqrt{\frac{\frac{2 \times 10^6}{3}}{600}} = \sqrt{\frac{2 \times 10^6}{1800}} = \sqrt{\frac{2000}{18}} = \sqrt{\frac{1000}{9}} = \frac{10}{3} \, \text{rad/s} \]

### Final Answer:
The angular frequency is:
\[ \omega = \frac{100}{3} \, \text{rad/s} \]