Question

Two identical strings of a stringed musical instrument are in unison when stretched with the same tension. When the tension in one string is increased by $1\%$, the musician hears 4 beats per second. What was the frequency of the note when the strings were in unison?

• A. 796 Hz
• B. 800 Hz
• C. 804 Hz
• D. 808 Hz

Answer 1

The correct option is B 800 Hz

Let $T_1$ be the tension in each string when they are in unison. Let $T_2$ be the increased tension such that $T_2=1.01T_1$.
Now, $\frac{f_2}{f_1}=\sqrt{\frac{T_2}{T_1}}$
$={(1+\frac{1}{100})}^{\frac{1}{2}}$
Expanding binomially, we have
${(1+\frac{1}{100})}^{\frac{1}{2}}=1+\frac{1}{200}=\frac{201}{200}$
Thus, $\frac{f_2}{f_1}=\frac{201}{200}$
Also, $f_2-f_1=4$
$\Rightarrow f_2=f_1+4$
Therefore, we have $\frac{f_1+4}{f_1}=\frac{201}{200}$
$\Rightarrow f_1=800Hz$
Hence, the correct choice is (b).