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.
If rod of length `4m`, area `4cm^(2)` amd young modules `2 xx 10^(10)N//m^(2)` is attached with mass `200 kg`, then angular frequency of SHM `("rad"//"sec.")` of mass is equal to -

Options

A. `1000`
B. `10`
C. `100`
D. `10 pi`

Answer

To solve the problem step by step, we will calculate the angular frequency of the simple harmonic motion (SHM) of a mass attached to a rod or cable treated as a spring.

### Step 1: Identify the given values
- Length of the rod, \( L = 4 \, \text{m} \)
- Area of the rod, \( A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \) (conversion to SI units)
- Young's modulus, \( Y = 2 \times 10^{10} \, \text{N/m}^2 \)
- Mass attached, \( m = 200 \, \text{kg} \)

### Step 2: Calculate the spring constant \( k \)
Using the formula for the spring constant derived from Young's modulus:
\[ k = \frac{AY}{L} \]
Substituting the values:
\[ k = \frac{(4 \times 10^{-4} \, \text{m}^2)(2 \times 10^{10} \, \text{N/m}^2)}{4 \, \text{m}} \]
Calculating the numerator:
\[ = 8 \times 10^{6} \, \text{N} \]
Now divide by the length:
\[ k = \frac{8 \times 10^{6} \, \text{N}}{4 \, \text{m}} = 2 \times 10^{6} \, \text{N/m} \]

### Step 3: Calculate the angular frequency \( \omega \)
The angular frequency for a mass-spring system is given by:
\[ \omega = \sqrt{\frac{k}{m}} \]
Substituting the values of \( k \) and \( m \):
\[ \omega = \sqrt{\frac{2 \times 10^{6} \, \text{N/m}}{200 \, \text{kg}}} \]
Calculating the fraction:
\[ = \sqrt{10^{4}} = 100 \, \text{rad/s} \]

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

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