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 solve the problem step by step, we will follow the outlined process to find the maximum transverse velocity amplitude of the vibrating string.

### Step 1: Determine the linear mass density (μ) of the string.
Given:
- Mass per unit length (μ) = 0.2 kg/m

### Step 2: Calculate the wavelength (λ) of the string.
The string vibrates in 3 segments, which means it has 3 half-wavelengths fitting into the length of the string.

- Length of the string (l) = 0.6 m
- Since the string vibrates in 3 segments, we can express this as:
\[ \frac{3}{2} \lambda = l \]
Rearranging gives:
\[ \lambda = \frac{2l}{3} = \frac{2 \times 0.6}{3} = 0.4 \text{ m} \]

### Step 3: Calculate the wave speed (v) on the string.
The wave speed on a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Where:
- T = tension = 80 N
- μ = linear mass density = 0.2 kg/m

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

### Step 4: Calculate the frequency (f) of the wave.
The frequency can be calculated using the wave speed and wavelength:
\[ f = \frac{v}{\lambda} \]
Substituting the values:
\[ f = \frac{20}{0.4} = 50 \text{ Hz} \]

### Step 5: Calculate the maximum transverse velocity amplitude (V_max).
The maximum transverse velocity amplitude is given by:
\[ V_{\text{max}} = \omega A \]
Where:
- \(\omega = 2\pi f\)
- A = maximum amplitude = 0.5 cm = 0.005 m

First, calculate \(\omega\):
\[ \omega = 2\pi f = 2 \times 3.14 \times 50 = 314 \text{ rad/s} \]

Now substitute \(\omega\) and A into the equation for \(V_{\text{max}}\):
\[ V_{\text{max}} = 314 \times 0.005 = 1.57 \text{ m/s} \]

### Final Answer:
The maximum transverse velocity amplitude is **1.57 m/s**.

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