Question

two violin strings are tuned to the same frequency, 294 hz the tension in one string is then decreased by 2.4%.
part a
what will be the beat frequency heard when the two strings are played together? hint recall the equation ( v=sqrt{\frac{f_{t}}{m / l}} ).
express your answer to two significant figures and include the appropriate units.
( f_{\text {beat }}= )

Explanation:

Step1: Recall the formula for the frequency of a vibrating string

The frequency of a vibrating string is given by \( f=\frac{n}{2L}\sqrt{\frac{F_T}{\mu}} \), where \( n \) is the number of harmonics, \( L \) is the length of the string, \( F_T \) is the tension and \( \mu \) is the linear mass density. For a given string ( \( n\), \( L\) and \( \mu\) constant), \( f\propto\sqrt{F_T}\). Let the initial frequency \( f_1 = 294\space Hz\) and the initial tension \( F_{T1}\), the new tension \( F_{T2}=(1 - 0.024)F_{T1}=0.976F_{T1}\).

Step2: Find the new frequency \( f_2\)

Since \( f\propto\sqrt{F_T}\), we have \( \frac{f_2}{f_1}=\sqrt{\frac{F_{T2}}{F_{T1}}}\). Substituting \( F_{T2} = 0.976F_{T1}\), we get \( f_2=f_1\sqrt{0.976}\).

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$$ f_2\approx290\space Hz$$

Step3: Calculate the beat frequency \( f_{beat}\)

The beat frequency formula is \( f_{beat}=\vert f_1 - f_2\vert\). Substituting \( f_1 = 294\space Hz\) and \( f_2\approx290\space Hz\), we get \( f_{beat}=294 - 290 = 4\space Hz\)

Answer:

\( f_{beat}=3.6\space Hz\)

Detailed calculation for more precision:

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\( f_{beat}=\vert294 - 290.26\vert=3.74\approx3.6\space Hz\) (rounded to two - significant figures)