A string in a musical instrument is 50 cm long and its fundamental frequency is 800 Hz. If a frequency of 1000 Hz is to be produced, then required length of string is

Options

A. 62.5 cm
B. 40 cm
C. 50 cm
D. 37.5 cm

Explanation

To solve the problem step by step, we will use the relationship between the frequency of a vibrating string and its length.

### Step 1: Understand the relationship between frequency and length
The fundamental frequency \( f \) of a string is inversely proportional to its length \( L \). This can be expressed mathematically as:
\[ f \propto \frac{1}{L} \]
This means that if the frequency increases, the length of the string must decrease, and vice versa.

### Step 2: Set up the known values
From the problem, we have:
- Initial length of the string, \( L_1 = 50 \, \text{cm} \)
- Initial frequency, \( f_1 = 800 \, \text{Hz} \)
- New frequency, \( f_2 = 1000 \, \text{Hz} \)

We need to find the new length \( L_2 \) when the frequency is \( f_2 \).

### Step 3: Use the proportionality relationship
From the relationship established, we can write:
\[ \frac{f_1}{f_2} = \frac{L_2}{L_1} \]

### Step 4: Substitute the known values
Substituting the known values into the equation:
\[ \frac{800 \, \text{Hz}}{1000 \, \text{Hz}} = \frac{L_2}{50 \, \text{cm}} \]

### Step 5: Simplify the equation
This simplifies to:
\[ \frac{8}{10} = \frac{L_2}{50} \]
or
\[ \frac{4}{5} = \frac{L_2}{50} \]

### Step 6: Solve for \( L_2 \)
Now, we can cross-multiply to solve for \( L_2 \):
\[ 4 \times 50 = 5 \times L_2 \]
\[ 200 = 5L_2 \]
\[ L_2 = \frac{200}{5} = 40 \, \text{cm} \]

### Conclusion
The required length of the string to produce a frequency of 1000 Hz is:
\[ \boxed{40 \, \text{cm}} \]