A string is stretched betweeb fixed points separated by `75.0 cm`. It observed to have resonant frequencies of `420 Hz` and `315 Hz`. There are no other resonant frequencies between these two. The lowest resonant frequency for this strings is

Answer

To find the lowest resonant frequency of a string stretched between two fixed points, we can follow these steps:

### Step 1: Understand the relationship between frequencies and harmonics
The resonant frequencies of a string fixed at both ends can be expressed as:
\[ f_n = n \frac{v}{2L} \]
where:
- \( f_n \) is the nth harmonic frequency,
- \( n \) is the harmonic number (1 for the fundamental frequency, 2 for the first overtone, etc.),
- \( v \) is the wave speed on the string,
- \( L \) is the length of the string.

### Step 2: Identify the given frequencies
We are given two resonant frequencies:
- \( f_1 = 315 \, \text{Hz} \)
- \( f_2 = 420 \, \text{Hz} \)

### Step 3: Set up equations for the frequencies
Assuming \( f_1 \) corresponds to the nth harmonic and \( f_2 \) corresponds to the (n+1)th harmonic, we can write:
1. \( 315 = n \frac{v}{2L} \) (Equation 1)
2. \( 420 = (n + 1) \frac{v}{2L} \) (Equation 2)

### Step 4: Divide the two equations
To eliminate \( v/2L \), we can divide Equation 2 by Equation 1:
\[ \frac{420}{315} = \frac{(n + 1)}{n} \]
This simplifies to:
\[ \frac{4}{3} = \frac{(n + 1)}{n} \]

### Step 5: Solve for \( n \)
Cross-multiply to solve for \( n \):
\[ 4n = 3(n + 1) \]
Expanding gives:
\[ 4n = 3n + 3 \]
Subtracting \( 3n \) from both sides:
\[ n = 3 \]

### Step 6: Substitute \( n \) back to find the fundamental frequency
Now that we have \( n = 3 \), we can substitute back into Equation 1 to find the fundamental frequency:
\[ 315 = 3 \frac{v}{2L} \]
Rearranging gives:
\[ \frac{v}{2L} = \frac{315}{3} = 105 \, \text{Hz} \]
Thus, the fundamental frequency (the lowest frequency) is:
\[ f_1 = 105 \, \text{Hz} \]

### Final Answer
The lowest resonant frequency for the string is:
\[ \boxed{105 \, \text{Hz}} \]