n number of waves are produced on a string in 0.5 s. Now, the tension in the string is doubled (Assume length and radius constant), the number of waves produced in 0.5s for the same harmonic will be

Options

A. n
B. `sqrt2n`
C. `n/sqrt2`
D. `n/sqrt5`

Answer

To solve the problem, we need to understand how the frequency of waves on a string is affected by the tension in the string. Here’s a step-by-step solution:

### Step 1: Understand the relationship between frequency, tension, and mass per unit length
The frequency \( f \) of a wave on a string is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) is the length of the string,
- \( T \) is the tension in the string,
- \( \mu \) is the mass per unit length of the string.

### Step 2: Determine the initial frequency
Initially, the frequency is given as \( n \) waves in 0.5 seconds. Therefore, the frequency \( f \) can be expressed as:
\[ f = \frac{n}{0.5} = 2n \text{ (waves per second)} \]

### Step 3: Analyze the effect of doubling the tension
Now, if the tension in the string is doubled, we denote the new tension as \( T' = 2T \). The new frequency \( f' \) can be expressed as:
\[ f' = \frac{1}{2L} \sqrt{\frac{T'}{\mu}} = \frac{1}{2L} \sqrt{\frac{2T}{\mu}} = \frac{1}{2L} \sqrt{2} \sqrt{\frac{T}{\mu}} \]

### Step 4: Relate the new frequency to the original frequency
We can express the new frequency \( f' \) in terms of the original frequency \( f \):
\[ f' = \sqrt{2} \cdot f \]
Substituting \( f = 2n \):
\[ f' = \sqrt{2} \cdot (2n) = 2\sqrt{2}n \]

### Step 5: Calculate the number of waves in 0.5 seconds with the new frequency
Now, we want to find the number of waves produced in 0.5 seconds with the new frequency \( f' \):
\[ \text{Number of waves} = f' \times 0.5 = (2\sqrt{2}n) \times 0.5 = \sqrt{2}n \]

### Conclusion
Thus, the number of waves produced in 0.5 seconds for the same harmonic after doubling the tension is:
\[ \sqrt{2}n \]