1. Beat frequency relation

For two nearby frequencies $f_1$ and $f_2$, the beat frequency is

$f_b = |f_1 - f_2|$

Here, the tuning fork frequency is

$f_t = 256\,\text{Hz}$

and initially the piano string makes $5$ beats per second, so

$|f_p - 256| = 5$

Thus the possible initial frequencies of the piano string are

$f_p = 256 + 5 = 261\,\text{Hz}$

or

$f_p = 256 - 5 = 251\,\text{Hz}$

2. Effect of increasing tension

For a string,

$f \propto \sqrt{T}$

So when the tension is increased, the frequency of the piano string increases.

After increasing tension, the beat frequency decreases from $5$ to $2$. That means the piano-string frequency moves closer to $256\,\text{Hz}$.

Now check both possibilities:

• If initially $f_p = 261\,\text{Hz}$, then increasing tension would increase it further above $261\,\text{Hz}$, making it move farther away from $256\,\text{Hz}$. The beat frequency would increase, not decrease. So this is impossible.

• If initially $f_p = 251\,\text{Hz}$, then increasing tension raises the frequency toward $256\,\text{Hz}$, so the beat frequency can decrease from $5$ to $2$. This is consistent.

Therefore, the initial frequency of the piano string was

$251\,\text{Hz} = 256 - 5\,\text{Hz}$

3. Option check

• A: $256 + 2 = 258\,\text{Hz}$ ❌
• B: $256 - 2 = 254\,\text{Hz}$ ❌
• C: $256 - 5 = 251\,\text{Hz}$ ✅
• D: $256 + 5 = 261\,\text{Hz}$ ❌

Hence, the correct option is C.