Question

Two identical piano strings of length 0.75m are each tuned exactly to 440Hz. The tension in one of the strings is then increased by 2%. If they are now struck, what is the beat frequency between the fundamental of the two strings?

• A. 4.3 Hz
• B. 15.3 Hz
• C. 7.3 Hz
• D. 18.3 Hz

Answer 1

The correct option is A 4.3 Hz

When the tension in one of the strings is changed, its fundamental frequancy also changes. Therefore, when both strings are played, they will have different frequencies and beats will be heard.
Use equation $v=\sqrt{\frac{T}{\mu }}$ to substitute for the wave speed on the strings.
$\frac{f_2}{f_2}=\frac{\sqrt{\frac{T_2}{\mu }}}{\sqrt{\frac{T_2}{\mu }}}=\sqrt{\frac{T_2}{T_1}}$
The tension in one string is 20% larger than the other, that is $T_2=1.010T_1$.

$\frac{f_2}{f_1}=\frac{\sqrt{1.02T_1}}{\sqrt{T_1}}=1.01$
Solve for the frequency of the tightened string:
$f_2=1.01f_1=1.01\left(440Hz\right)=444.3Hz$

$f_{beat}=444.3Hz-440Hz=4.3Hz$