A piano wire `0.5m` long and mass `5gm` is streteched by a tension of `400N` .The number of highest overtone that can be heared by a person is

Options

A. `160`
B. `99`
C. `140`
D. `120`

Explanation

To solve the problem, we need to find the number of highest overtones that can be heard by a person when a piano wire is stretched under a certain tension. We will use the formula for the frequency of the vibrating string and the relationship between harmonics and overtones.

### Step-by-Step Solution:

1. **Convert Mass to Kilograms:**
The mass of the piano wire is given as 5 grams. We need to convert this to kilograms.
\[ \text{Mass} (m) = 5 \text{ grams} = 5 \times 10^{-3} \text{ kg} \]

**Hint:** Remember that 1 gram = \(10^{-3}\) kg.

2. **Calculate the Linear Mass Density (μ):**
The linear mass density (μ) is defined as mass per unit length.
\[ \mu = \frac{m}{L} = \frac{5 \times 10^{-3} \text{ kg}}{0.5 \text{ m}} = 0.01 \text{ kg/m} \]

**Hint:** Linear mass density is calculated by dividing the mass of the wire by its length.

3. **Use the Formula for Frequency of the Fundamental Mode:**
The frequency of the fundamental mode (first harmonic) for a stretched string is given by:
\[ f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \(T\) is the tension in the wire and \(L\) is the length of the wire.

Substituting the values:
\[ f_1 = \frac{1}{2 \times 0.5} \sqrt{\frac{400 \text{ N}}{0.01 \text{ kg/m}}} \]

**Hint:** The fundamental frequency is calculated using the tension and linear mass density.

4. **Calculate the Value Inside the Square Root:**
\[ f_1 = \frac{1}{1} \sqrt{40000} = \sqrt{40000} = 200 \text{ Hz} \]

**Hint:** Simplify the square root to find the frequency.

5. **Calculate the Frequencies of Higher Harmonics:**
The frequency of the nth harmonic is given by:
\[ f_n = n \cdot f_1 \]
where \(n\) is the harmonic number.

6. **Determine the Highest Overtone:**
The highest overtone corresponds to the highest harmonic. The number of overtones is given by \(n - 1\) for the nth harmonic. The maximum audible frequency for a human is typically around 20 kHz.

Set \(f_n \leq 20000 \text{ Hz}\):
\[ n \cdot 200 \leq 20000 \]
\[ n \leq \frac{20000}{200} = 100 \]

Therefore, the highest overtone \(n - 1\) is:
\[ \text{Highest overtone} = 100 - 1 = 99 \]

**Hint:** The highest overtone is one less than the maximum harmonic number that can be heard.

### Final Answer:
The highest overtone that can be heard by a person is **99**.

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