Four standing wave segments, or loops, are observed on a string fixed at both ends as it vibrates at a frequency of 140 Hz. What is the fundamental frequency of the string?

Options

A. 23 Hz
B. 28 Hz
C. 35 Hz
D. 47 Hz

Answer

To solve the problem, we need to find the fundamental frequency of a string that is vibrating in the fourth harmonic (4 loops) at a frequency of 140 Hz. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Harmonics
The string is fixed at both ends, and it is vibrating in the fourth harmonic. The number of loops (or segments) corresponds to the harmonic number (n). In this case, n = 4.

### Step 2: Write the Equation for Harmonics
For a string fixed at both ends, the relationship between the harmonic number (n), the wavelength (λ), and the length of the string (L) is given by:
\[ n \frac{\lambda}{2} = L \]
For the fourth harmonic (n = 4), we can write:
\[ 4 \frac{\lambda}{2} = L \]
This simplifies to:
\[ 2\lambda = L \]
Thus, the wavelength is:
\[ \lambda = \frac{L}{2} \]

### Step 3: Relate Wave Velocity, Frequency, and Wavelength
The wave velocity (v) on the string can be expressed in terms of frequency (f) and wavelength (λ):
\[ v = f \lambda \]
For the fourth harmonic, substituting for λ gives:
\[ v = 140 \, \text{Hz} \cdot \lambda \]

### Step 4: Find the Velocity
From the equation for the fourth harmonic:
\[ \lambda = \frac{L}{2} \]
We can substitute this into the wave velocity equation:
\[ v = 140 \cdot \frac{L}{2} \]

### Step 5: Set Up the Fundamental Frequency Equation
Now, we need to find the fundamental frequency (f₁) when the string vibrates in the first harmonic (n = 1). The relationship for the fundamental frequency is:
\[ 1 \frac{\lambda_1}{2} = L \]
This means:
\[ \lambda_1 = 2L \]
Using the wave velocity equation:
\[ v = f_1 \lambda_1 \]
Substituting for λ₁ gives:
\[ v = f_1 \cdot 2L \]

### Step 6: Equate the Two Expressions for Velocity
Now we have two expressions for the wave velocity:
1. From the fourth harmonic:
\[ v = 140 \cdot \frac{L}{2} \]
2. From the fundamental frequency:
\[ v = f_1 \cdot 2L \]

Setting these equal to each other:
\[ 140 \cdot \frac{L}{2} = f_1 \cdot 2L \]

### Step 7: Solve for the Fundamental Frequency
We can cancel L from both sides (assuming L ≠ 0):
\[ 140 \cdot \frac{1}{2} = f_1 \cdot 2 \]
This simplifies to:
\[ 70 = 2f_1 \]
Dividing both sides by 2 gives:
\[ f_1 = 35 \, \text{Hz} \]

### Conclusion
The fundamental frequency of the string is **35 Hz**.