A mass of 0.2 kg, length 1m and cross sectional area `4.9 xx 10^(-7) m^(2)` is suspended from a massless wire. The mass is pulled slightly in the vertically downward direction and released. It performs simple harmonic motion of angular frequency 70 rad `s^(-1)`. If the Young's modulus of the material of the wire is `x xx 10^(9) Nm^(-2)`, find the value of x.

Explanation

To find the value of \( x \) in the Young's modulus of the material of the wire, we can follow these steps:

### Step 1: Understand the relationship between Young's modulus, stress, and strain
Young's modulus \( Y \) is defined as the ratio of stress to strain:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]

### Step 2: Define stress and strain
Stress (\( \sigma \)) is given by:
\[ \sigma = \frac{F}{A} \]
where \( F \) is the force applied, and \( A \) is the cross-sectional area.

Strain (\( \epsilon \)) is defined as:
\[ \epsilon = \frac{\Delta L}{L} \]
where \( \Delta L \) is the change in length (extension) and \( L \) is the original length.

### Step 3: Calculate the force acting on the wire
The force acting on the wire due to the suspended mass is the weight of the mass:
\[ F = mg \]
where \( m = 0.2 \, \text{kg} \) and \( g \approx 9.81 \, \text{m/s}^2 \). Thus,
\[ F = 0.2 \times 9.81 = 1.962 \, \text{N} \]

### Step 4: Substitute values into the stress formula
The cross-sectional area \( A \) is given as \( 4.9 \times 10^{-7} \, \text{m}^2 \). Therefore, the stress is:
\[ \sigma = \frac{F}{A} = \frac{1.962}{4.9 \times 10^{-7}} \approx 4.0 \times 10^{6} \, \text{N/m}^2 \]

### Step 5: Relate angular frequency to spring constant
The angular frequency \( \omega \) of the simple harmonic motion is given as \( 70 \, \text{rad/s} \). The spring constant \( k \) can be related to \( \omega \) and mass \( m \) by:
\[ \omega = \sqrt{\frac{k}{m}} \implies k = m\omega^2 \]
Substituting the values:
\[ k = 0.2 \times (70)^2 = 0.2 \times 4900 = 980 \, \text{N/m} \]

### Step 6: Calculate the strain
Using the relationship \( \Delta L = \frac{mg}{k} \):
\[ \Delta L = \frac{1.962}{980} \approx 0.002 \, \text{m} \]
Now, the strain is:
\[ \epsilon = \frac{\Delta L}{L} = \frac{0.002}{1} = 0.002 \]

### Step 7: Substitute into Young's modulus formula
Now substituting stress and strain into the Young's modulus formula:
\[ Y = \frac{\sigma}{\epsilon} = \frac{4.0 \times 10^{6}}{0.002} = 2.0 \times 10^{9} \, \text{N/m}^2 \]

### Step 8: Find \( x \)
Given that \( Y = x \times 10^{9} \, \text{N/m}^2 \), we can equate:
\[ 2.0 \times 10^{9} = x \times 10^{9} \]
Thus, \( x = 2 \).

### Final Answer
The value of \( x \) is \( 2 \).

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