A string of mass 0.2 kg/m and length l= 0.6 m is fixed at both ends and stretched such that it has a tension of 80 N. The string vibrates in 3 segments with maximum amplitude of 0.5 cm. the maximum transverse velocity amplitude is

Options

A. 1.57 m/s
B. 6. 28 m/s
C. 3.14 m/s
D. 9.42 m/s

Answer

To find the maximum transverse velocity amplitude of the vibrating string, we can follow these steps:

### Step 1: Identify the given values
- Mass per unit length of the string (μ) = 0.2 kg/m
- Length of the string (L) = 0.6 m
- Tension in the string (T) = 80 N
- Number of segments (n) = 3
- Maximum amplitude (A) = 0.5 cm = 0.005 m

### Step 2: Calculate the wavelength (λ)
Since the string vibrates in 3 segments, the length of the string is given by:
\[ L = \frac{3\lambda}{2} \]

Rearranging this gives:
\[ \lambda = \frac{2L}{3} = \frac{2 \times 0.6}{3} = 0.4 \, \text{m} \]

### Step 3: Calculate the wave velocity (v)
The wave velocity can be calculated using the formula:
\[ v = \sqrt{\frac{T}{μ}} \]

Substituting the values:
\[ v = \sqrt{\frac{80}{0.2}} = \sqrt{400} = 20 \, \text{m/s} \]

### Step 4: Calculate the frequency (f)
Using the relationship between wave velocity, frequency, and wavelength:
\[ v = f \lambda \]

Rearranging gives:
\[ f = \frac{v}{\lambda} = \frac{20}{0.4} = 50 \, \text{Hz} \]

### Step 5: Calculate the angular frequency (ω)
The angular frequency is given by:
\[ \omega = 2\pi f \]

Substituting the value of frequency:
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

### Step 6: Calculate the maximum transverse velocity amplitude (VA)
The maximum transverse velocity amplitude is given by:
\[ V_A = \omega A \]

Substituting the values:
\[ V_A = 100\pi \times 0.005 = 0.5\pi \, \text{m/s} \]

### Step 7: Calculate the numerical value
Using \( \pi \approx 3.14 \):
\[ V_A \approx 0.5 \times 3.14 \approx 1.57 \, \text{m/s} \]

### Final Answer
The maximum transverse velocity amplitude is approximately:
\[ V_A \approx 1.57 \, \text{m/s} \]

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