Step 1: Understanding the Concept:
The fundamental frequency of a vibrating string depends inversely on its length, provided the tension and linear mass density are kept constant.
When the length is increased, the frequency will decrease. The percentage change in the fundamental frequency will be the same as the percentage change in any of its harmonics (like the second harmonic).
Step 2: Key Formula or Approach:
The fundamental frequency $f$ of a sonometer wire is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.
From this, $f \propto \frac{1}{L}$.
The frequency of the $n$-th harmonic is $f_n = n \times f$.
Step 3: Detailed Explanation:
Let the initial length be $L_1 = L$ and the initial fundamental frequency be $f_1 = 50 \text{ Hz}$.
The new length $L_2$ is increased by 25%:
\[ L_2 = L + 0.25L = 1.25L = \frac{5}{4}L \]
Since $f \propto \frac{1}{L}$, the new fundamental frequency $f_2$ is:
\[ \frac{f_2}{f_1} = \frac{L_1}{L_2} = \frac{L}{\frac{5}{4}L} = \frac{4}{5} \]
\[ f_2 = \frac{4}{5}f_1 \]
The fractional change in fundamental frequency is:
\[ \frac{\Delta f}{f_1} = \frac{f_2 - f_1}{f_1} = \frac{\frac{4}{5}f_1 - f_1}{f_1} = -\frac{1}{5} \]
Percentage change $= -\frac{1}{5} \times 100% = -20%$.
The frequency of the second harmonic is $2f$. Its new frequency will be $2f_2 = 2(\frac{4}{5}f_1) = \frac{4}{5}(2f_1)$.
The percentage change for the second harmonic is identical to the fundamental frequency's percentage change, which is a $20%$ decrease.
Step 4: Final Answer:
The frequency change of the second harmonic is decreased by 20%.