A `0.1 kg` mass is suspended from a wire of negligible mass. The length of the wire is `1 m` and its cross - sectional area is `4.9xx10^(-7) m^(2)`. If the mass is pulled a little in the vertically downward direction and released , it performs `SHM` with angular frequency `140 rad s^(-1)`. If the young's modulus of the material of the wire is ` pxx10^(9) Nm^(-2)`, find the value of `p`.

Explanation

To solve the problem, we will use the formula for the angular frequency \( \omega \) of a mass-spring system, which can be related to Young's modulus \( Y \) of the wire. The formula is given by:

\[ \omega = \sqrt{\frac{Y \cdot A}{M \cdot L}} \]

Where:
- \( \omega \) is the angular frequency,
- \( Y \) is Young's modulus,
- \( A \) is the cross-sectional area of the wire,
- \( M \) is the mass suspended,
- \( L \) is the length of the wire.

### Step 1: Identify the known values
From the problem, we have:
- Mass \( M = 0.1 \, \text{kg} \)
- Length \( L = 1 \, \text{m} \)
- Cross-sectional area \( A = 4.9 \times 10^{-7} \, \text{m}^2 \)
- Angular frequency \( \omega = 140 \, \text{rad/s} \)

### Step 2: Substitute the known values into the formula
We can rearrange the formula to solve for Young's modulus \( Y \):

\[ Y = \frac{\omega^2 \cdot M \cdot L}{A} \]

Substituting the known values:

\[ Y = \frac{(140)^2 \cdot (0.1) \cdot (1)}{4.9 \times 10^{-7}} \]

### Step 3: Calculate \( \omega^2 \)
First, calculate \( \omega^2 \):

\[ \omega^2 = 140^2 = 19600 \, \text{(rad/s)}^2 \]

### Step 4: Substitute and calculate \( Y \)
Now substitute \( \omega^2 \) back into the equation for \( Y \):

\[ Y = \frac{19600 \cdot 0.1 \cdot 1}{4.9 \times 10^{-7}} \]

\[ Y = \frac{1960}{4.9 \times 10^{-7}} \]

### Step 5: Perform the division
Now calculate the division:

\[ Y = 1960 \div (4.9 \times 10^{-7}) = 1960 \times \frac{1}{4.9} \times 10^{7} \]

Calculating \( \frac{1960}{4.9} \):

\[ \frac{1960}{4.9} \approx 400 \]

Thus,

\[ Y \approx 400 \times 10^{7} \, \text{N/m}^2 = 4 \times 10^{9} \, \text{N/m}^2 \]

### Step 6: Identify the value of \( p \)
From the problem, we have \( Y = p \times 10^{9} \, \text{N/m}^2 \). Therefore, we find:

\[ p = 4 \]

### Final Answer
The value of \( p \) is \( 4 \).

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