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 1: Identify the wave equation
The given wave equation is:
\[ y = 2 \sin(0.3 \, \text{cm}^{-1}) x \cos(500 \pi \, \text{s}^{-1} t) \, \text{cm} \]

### Step 2: Compare with the standard wave equation
The standard form of the wave equation is:
\[ y = A \sin(kx) \cos(\omega t) \]
where:
- \( A \) is the amplitude,
- \( k \) is the wave number,
- \( \omega \) is the angular frequency.

From the given equation, we can identify:
- \( k = 0.3 \, \text{cm}^{-1} \)

### Step 3: Relate wave number to wavelength
The wave number \( k \) is related to the wavelength \( \lambda \) by the formula:
\[ k = \frac{2\pi}{\lambda} \]
We can rearrange this to find \( \lambda \):
\[ \lambda = \frac{2\pi}{k} \]

### Step 4: Calculate the wavelength
Substituting the value of \( k \):
\[ \lambda = \frac{2\pi}{0.3} \]
Calculating this gives:
\[ \lambda = \frac{2 \times 3.14}{0.3} = \frac{6.28}{0.3} = 20.93 \, \text{cm} \]

### Step 5: Determine the length of the string in the second harmonic
For a string fixed at both ends, the length \( L \) in the \( n \)-th harmonic is given by:
\[ L = \frac{n \lambda}{2} \]
For the second harmonic (\( n = 2 \)):
\[ L = \frac{2 \lambda}{2} = \lambda \]

### Step 6: Substitute the wavelength to find the length
Thus, the length of the string is:
\[ L = 20.93 \, \text{cm} \]

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

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