A sting of length 1m and linear mass density 0.01 `kg//m` is stretched to a tension of 100 N. When both ends of the string are fixed, the three lowest frequencies for standing wave are `f_1, f_2 and f_3` . When only one end of the string is fixed, the three lowest frequencies for standing wave are `n_1, n_2 and n_3`. Then,

Options

A. `n_3 = 5n_1 = f_3 = 125 Hz`
B. `f_3 = 5f_1 = n_2 = 125 Hz`
C. `f_3 = n_2 = 3f_1 = 150 Hz `
D. `n_2 = (f_1 + f_2)/2 = 75 Hz`

Explanation

To solve the problem, we need to find the three lowest frequencies for standing waves in two different scenarios: when both ends of the string are fixed and when one end is fixed.

### Step 1: Calculate the speed of the wave on the string
The speed of a wave on a string can be calculated using the formula:

\[ v = \sqrt{\frac{T}{\mu}} \]

where:
- \( T = 100 \, \text{N} \) (tension)
- \( \mu = 0.01 \, \text{kg/m} \) (linear mass density)

Substituting the values:

\[ v = \sqrt{\frac{100}{0.01}} = \sqrt{10000} = 100 \, \text{m/s} \]

### Step 2: Calculate the frequencies when both ends are fixed
When both ends of the string are fixed, the standing wave frequencies can be calculated using the formula:

\[ f_n = \frac{n \cdot v}{2L} \]

where \( n \) is the harmonic number, \( L = 1 \, \text{m} \) (length of the string).

- **Fundamental frequency (n = 1)**:
\[ f_1 = \frac{1 \cdot 100}{2 \cdot 1} = 50 \, \text{Hz} \]

- **Second harmonic (n = 2)**:
\[ f_2 = \frac{2 \cdot 100}{2 \cdot 1} = 100 \, \text{Hz} \]

- **Third harmonic (n = 3)**:
\[ f_3 = \frac{3 \cdot 100}{2 \cdot 1} = 150 \, \text{Hz} \]

Thus, the frequencies when both ends are fixed are:
- \( f_1 = 50 \, \text{Hz} \)
- \( f_2 = 100 \, \text{Hz} \)
- \( f_3 = 150 \, \text{Hz} \)

### Step 3: Calculate the frequencies when one end is fixed
When one end of the string is fixed, the standing wave frequencies are given by:

\[ n_k = \frac{k \cdot v}{4L} \]

where \( k \) is the harmonic number (only odd harmonics are present).

- **Fundamental frequency (k = 1)**:
\[ n_1 = \frac{1 \cdot 100}{4 \cdot 1} = 25 \, \text{Hz} \]

- **Third harmonic (k = 3)**:
\[ n_2 = \frac{3 \cdot 100}{4 \cdot 1} = 75 \, \text{Hz} \]

- **Fifth harmonic (k = 5)**:
\[ n_3 = \frac{5 \cdot 100}{4 \cdot 1} = 125 \, \text{Hz} \]

Thus, the frequencies when one end is fixed are:
- \( n_1 = 25 \, \text{Hz} \)
- \( n_2 = 75 \, \text{Hz} \)
- \( n_3 = 125 \, \text{Hz} \)

### Step 4: Compare the frequencies
Now, we can compare the frequencies from both scenarios:
- \( f_1 : f_2 : f_3 = 50 : 100 : 150 \)
- \( n_1 : n_2 : n_3 = 25 : 75 : 125 \)

### Step 5: Analyze the relationships
We can express the relationships:
- \( f_3 = 3 \cdot f_1 \)
- \( n_2 = \frac{f_1 + f_2}{2} \)
- \( n_3 = 5 \cdot n_1 \) is false as \( n_3 = 125 \) is not \( 5 \cdot 25 \).

### Conclusion
The correct relationships are:
- \( f_3 = 3 \cdot f_1 \)
- \( n_2 = 75 \) is equal to \( \frac{f_1 + f_2}{2} \).