The equation for the vibration of a string fixed at both ends vibrating in its third harmonic is given by `y=2cm sin[(0.6cm^-1)x]cos[(500pis^-1)t]`. The length of the string is

Options

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

Explanation

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

### Step 1: Identify the wave equation
The given wave equation is:
\[ y = 2 \, \text{cm} \sin(0.6 \, \text{cm}^{-1} \, x) \cos(500 \pi \, \text{s}^{-1} \, t) \]

### Step 2: Identify the wave number (k)
From the equation, we can see that:
\[ k = 0.6 \, \text{cm}^{-1} \]

### Step 3: Relate wave number to wavelength (λ)
The wave number \( k \) is related to the wavelength \( \lambda \) by the formula:
\[ k = \frac{2\pi}{\lambda} \]
Thus, we can rearrange this to find \( \lambda \):
\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{0.6} \]

### Step 4: Calculate the wavelength (λ)
Now, we can calculate \( \lambda \):
\[ \lambda = \frac{2\pi}{0.6} \approx \frac{6.2832}{0.6} \approx 10.47 \, \text{cm} \]

### Step 5: Determine the length of the string in the third harmonic
For a string fixed at both ends vibrating in its \( n \)-th harmonic, the length \( L \) of the string is given by:
\[ L = \frac{n}{2} \lambda \]
For the third harmonic (\( n = 3 \)):
\[ L = \frac{3}{2} \lambda = \frac{3}{2} \times 10.47 \, \text{cm} \]

### Step 6: Calculate the length of the string (L)
Now, we can calculate \( L \):
\[ L = \frac{3}{2} \times 10.47 \approx 15.705 \, \text{cm} \]

### Final Answer
The length of the string is approximately:
\[ L \approx 15.7 \, \text{cm} \]

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