The equation for the vibration of a string fixed at both ends vibrating in its second harmonic is given by `y=2sin(0.3cm^(-1))xcos((500pis^(-1))t)cm`. The length of the string is :

Options

A. 24.6cm
B. 12.5cm
C. 20.6 cm
D. 15.7 cm

Explanation

To find the length of the string vibrating in its second harmonic, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given equation**: The equation for the vibration of the string is given by:
\[ y = 2 \sin(0.3 \, \text{cm}^{-1}) x \cos(500 \pi \, \text{s}^{-1} t) \, \text{cm} \]

2. **Compare with the general form**: The general form of the wave equation is:
\[ y = A \sin(kx) \cos(\omega t) \]
From the given equation, we can identify:
- Amplitude \( A = 2 \, \text{cm} \)
- Wave number \( k = 0.3 \, \text{cm}^{-1} \)
- Angular frequency \( \omega = 500 \pi \, \text{s}^{-1} \)

3. **Calculate the wavelength**: The wave number \( k \) is related to the wavelength \( \lambda \) by the formula:
\[ k = \frac{2\pi}{\lambda} \]
Rearranging this gives:
\[ \lambda = \frac{2\pi}{k} \]
Substituting the value of \( k \):
\[ \lambda = \frac{2\pi}{0.3} = \frac{20\pi}{3} \, \text{cm} \]

4. **Determine the length of the string**: For a string fixed at both ends vibrating in its second harmonic, the length \( L \) is given by:
\[ L = \frac{n\lambda}{2} \]
where \( n \) is the harmonic number. For the second harmonic, \( n = 2 \):
\[ L = \frac{2 \cdot \lambda}{2} = \lambda \]
Thus, substituting the value of \( \lambda \):
\[ L = \frac{20\pi}{3} \, \text{cm} \]

5. **Calculate the numerical value**: Using \( \pi \approx 3.14 \):
\[ L \approx \frac{20 \cdot 3.14}{3} \approx \frac{62.8}{3} \approx 20.93 \, \text{cm} \]

### Final Answer:
The length of the string is approximately \( 20.93 \, \text{cm} \).

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