A person hums in a well and finds strong resonance at frequencies `60 Hz, 100Hz` and `140 Hz`. What is the fundamental frequency of the well? Explain? How deep is the wall ? (velocity of sound `= 344 m//`).

Explanation

To solve the problem step by step, we will follow these procedures:

### Step 1: Understand the Problem
We are given the frequencies at which resonance occurs in a well: 60 Hz, 100 Hz, and 140 Hz. We need to find the fundamental frequency of the well and its depth.

### Step 2: Identify the Harmonics
The frequencies given correspond to harmonics of a pipe that is closed at one end (the well). For a pipe closed at one end, the harmonics are given by:
- First harmonic (fundamental frequency, \( f_0 \))
- Third harmonic (\( f_1 = 3f_0 \))
- Fifth harmonic (\( f_2 = 5f_0 \))
- Seventh harmonic (\( f_3 = 7f_0 \))

Given the frequencies:
- \( f_1 = 60 \, \text{Hz} \) (third harmonic)
- \( f_2 = 100 \, \text{Hz} \) (fifth harmonic)
- \( f_3 = 140 \, \text{Hz} \) (seventh harmonic)

### Step 3: Establish Ratios
From the harmonics, we can establish the following ratios:
- \( f_1 : f_2 : f_3 = 60 : 100 : 140 \)
- Simplifying this gives us \( 3 : 5 : 7 \).

### Step 4: Calculate the Fundamental Frequency
Using the ratios, we can express the fundamental frequency \( f_0 \):
- From the third harmonic: \( f_1 = 3f_0 \)
- So, \( f_0 = \frac{60}{3} = 20 \, \text{Hz} \)

### Step 5: Calculate the Depth of the Well
The fundamental frequency for a pipe closed at one end is given by the formula:
\[ f_0 = \frac{v}{4L} \]
where \( v \) is the velocity of sound and \( L \) is the length (depth) of the well. Rearranging this gives:
\[ L = \frac{v}{4f_0} \]

Substituting the values:
- \( v = 344 \, \text{m/s} \)
- \( f_0 = 20 \, \text{Hz} \)

Calculating \( L \):
\[ L = \frac{344}{4 \times 20} = \frac{344}{80} = 4.3 \, \text{m} \]

### Final Answers
- The fundamental frequency of the well is \( 20 \, \text{Hz} \).
- The depth of the well is \( 4.3 \, \text{m} \).

---