the equation for the vibration of a string fixed both ends vibration in its third harmonic is given by
y = 2 cm sin `[(0.6cm ^(-1)x)xx ] cos [(500 ps^(-1)t]`What is the position of node?

Options

A. 24.6 cm
B. 12.5 cm
C. 20 . 6 cm
D. 15. 7 cm

Answer

To find the position of nodes in the vibration of a string fixed at both ends in its third harmonic, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Wave Equation**:
The wave equation is given as:
\[ y = 2 \, \text{cm} \sin(0.6 \, \text{cm}^{-1} \, x) \cos(500 \, \text{ps}^{-1} \, t) \]
Here, \( k = 0.6 \, \text{cm}^{-1} \) and \( \omega = 500 \, \text{ps}^{-1} \).

2. **Calculate the Wavelength (\( \lambda \))**:
The wavelength can be calculated using the formula:
\[ \lambda = \frac{2\pi}{k} \]
Substituting the value of \( k \):
\[ \lambda = \frac{2\pi}{0.6} \approx \frac{6.2832}{0.6} \approx 10.47 \, \text{cm} \]

3. **Determine the Positions of Nodes**:
The positions of nodes in a standing wave can be found using the formula:
\[ x_n = \frac{n\lambda}{2} \]
where \( n \) is the node number (1, 2, 3,...).

- For the **1st node** (\( n = 1 \)):
\[ x_1 = \frac{1 \times 10.47}{2} \approx 5.24 \, \text{cm} \]

- For the **2nd node** (\( n = 2 \)):
\[ x_2 = \frac{2 \times 10.47}{2} \approx 10.47 \, \text{cm} \]

- For the **3rd node** (\( n = 3 \)):
\[ x_3 = \frac{3 \times 10.47}{2} \approx 15.71 \, \text{cm} \]

4. **Conclusion**:
The positions of the nodes for the third harmonic are:
- 1st node: \( 5.24 \, \text{cm} \)
- 2nd node: \( 10.47 \, \text{cm} \)
- 3rd node: \( 15.71 \, \text{cm} \)

### Summary:
The position of the nodes in the string fixed at both ends in its third harmonic is at approximately \( 5.24 \, \text{cm} \), \( 10.47 \, \text{cm} \), and \( 15.71 \, \text{cm} \).