Vibrations with frequency `6.00 xx 10^2` Hz are established on a 1.33 m length of string that is clamed at both ends. The speed of waves on the string is `4.0 xx 10^2 m//s`
How far from either end of the string does the first node occur?

Options

A. 0.17m
B. 0.49 m
C. 0.33m
D. 0.66 m

Answer

To solve the problem, we need to determine the distance from either end of the string to the first node of the standing wave. Here’s the step-by-step solution:

### Step 1: Understand the Problem
We have a string of length \( L = 1.33 \, \text{m} \) that is clamped at both ends. The frequency of the vibrations is \( f = 6.00 \times 10^2 \, \text{Hz} \) and the speed of the wave on the string is \( v = 4.0 \times 10^2 \, \text{m/s} \). We need to find the distance from either end of the string to the first node.

### Step 2: Determine the Wavelength
The relationship between wave speed, frequency, and wavelength is given by the formula:
\[ v = f \lambda \]
where \( \lambda \) is the wavelength. Rearranging this gives:
\[ \lambda = \frac{v}{f} \]
Substituting the given values:
\[ \lambda = \frac{400 \, \text{m/s}}{600 \, \text{Hz}} = \frac{400}{600} = \frac{2}{3} \, \text{m} \approx 0.667 \, \text{m} \]

### Step 3: Identify the Harmonic Mode
For a string clamped at both ends, the standing wave pattern forms nodes at both ends. The first node occurs at \( \frac{\lambda}{2} \) from either end of the string.

### Step 4: Calculate the Distance to the First Node
The distance from either end to the first node is:
\[ \text{Distance to first node} = \frac{\lambda}{2} = \frac{0.667 \, \text{m}}{2} \approx 0.3335 \, \text{m} \]

### Step 5: Final Answer
Thus, the distance from either end of the string to the first node is approximately:
\[ \text{Distance} \approx 0.33 \, \text{m} \]

### Summary
The first node occurs approximately \( 0.33 \, \text{m} \) from either end of the string.

---