A 4.00-m long string, clamped at both ends, vibrates at `2.00 xx 10^2` Hz. If the string resonates in six segments, what is the speed of transverse waves on the string?

Options

A. 100 m/s
B. 133 m/s
C. 267 m/s
D. 328 m/s

Answer

To find the speed of transverse waves on a string that is clamped at both ends and vibrating in six segments, we can follow these steps:

### Step 1: Understand the relationship between the length of the string, the wavelength, and the number of segments (harmonics).
The formula for the length of the string in terms of the wavelength (λ) and the number of segments (n) is given by:
\[ L = \frac{n \lambda}{2} \]
where \(L\) is the length of the string, \(n\) is the number of segments (or loops), and \(\lambda\) is the wavelength.

### Step 2: Substitute the known values into the equation.
Given that the length of the string \(L = 4.00 \, \text{m}\) and the number of segments \(n = 6\), we can substitute these values into the equation:
\[ 4.00 = \frac{6 \lambda}{2} \]
This simplifies to:
\[ 4.00 = 3 \lambda \]

### Step 3: Solve for the wavelength (λ).
To find the wavelength, we rearrange the equation:
\[ \lambda = \frac{4.00}{3} \approx 1.33 \, \text{m} \]

### Step 4: Use the wave speed formula.
The speed of a wave (v) is related to its frequency (f) and wavelength (λ) by the equation:
\[ v = f \lambda \]
We are given the frequency \(f = 2.00 \times 10^2 \, \text{Hz} = 200 \, \text{Hz}\).

### Step 5: Substitute the values into the wave speed formula.
Now, we can substitute the frequency and the wavelength into the wave speed equation:
\[ v = 200 \, \text{Hz} \times 1.33 \, \text{m} \]

### Step 6: Calculate the speed of the wave.
Calculating the above gives:
\[ v \approx 266.67 \, \text{m/s} \]
Rounding this, we find:
\[ v \approx 267 \, \text{m/s} \]

### Final Answer:
The speed of transverse waves on the string is approximately **267 m/s**.

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