The velocity of sound in air is `340ms^(-1)`. A pipe closed at one end has a length of 170 cm. neglecting the end correction, the frquencies at which the pipe can resonate are

Options

A. 50 Hz
B. 100 Hz
C. 150 hz
D. 200 Hz

Answer

To solve the problem of finding the frequencies at which a pipe closed at one end can resonate, we will use the formula for the frequencies of a closed pipe. Here are the steps to derive the solution:

### Step 1: Understand the Formula
For a pipe closed at one end, the resonant frequencies can be calculated using the formula:
\[ f_n = (2n + 1) \frac{v}{4L} \]
where:
- \( f_n \) = frequency of the nth harmonic
- \( n \) = harmonic number (0, 1, 2, ...)
- \( v \) = velocity of sound in air (340 m/s)
- \( L \) = length of the pipe in meters

### Step 2: Convert Length to Meters
The length of the pipe is given as 170 cm. We need to convert this to meters:
\[ L = 170 \, \text{cm} = \frac{170}{100} \, \text{m} = 1.7 \, \text{m} \]

### Step 3: Calculate the Fundamental Frequency (n = 0)
Using \( n = 0 \):
\[ f_0 = (2(0) + 1) \frac{340}{4 \times 1.7} \]
\[ f_0 = 1 \cdot \frac{340}{6.8} \]
\[ f_0 = \frac{340}{6.8} = 50 \, \text{Hz} \]

### Step 4: Calculate the First Overtone (n = 1)
Using \( n = 1 \):
\[ f_1 = (2(1) + 1) \frac{340}{4 \times 1.7} \]
\[ f_1 = 3 \cdot \frac{340}{6.8} \]
\[ f_1 = \frac{1020}{6.8} = 150 \, \text{Hz} \]

### Step 5: Calculate the Second Overtone (n = 2)
Using \( n = 2 \):
\[ f_2 = (2(2) + 1) \frac{340}{4 \times 1.7} \]
\[ f_2 = 5 \cdot \frac{340}{6.8} \]
\[ f_2 = \frac{1700}{6.8} = 250 \, \text{Hz} \]

### Step 6: List the Frequencies
The frequencies at which the pipe can resonate are:
- Fundamental frequency: \( 50 \, \text{Hz} \)
- First overtone: \( 150 \, \text{Hz} \)
- Second overtone: \( 250 \, \text{Hz} \)

### Conclusion
The frequencies at which the pipe can resonate are \( 50 \, \text{Hz} \), \( 150 \, \text{Hz} \), and \( 250 \, \text{Hz} \).

---