Question

two violin strings are tuned to the same frequency, 294 hz. the tension in one string is then decreased by 3.6%. part a what will be the beat frequency heard when the two strings are played together? hint: recall the equation $v = sqrt{\frac{t}{m/l}}$ express your answer to two significant figures and include the appropriate units. $f_{beat}=$value units submit request answer provide feedback

Explanation:

Step1: Recall wave - speed and frequency relation

The frequency of a vibrating string is $f=\frac{v}{\lambda}$, and $v = \sqrt{\frac{T}{\mu}}$ where $T$ is tension and $\mu=\frac{m}{L}$ is linear mass - density. Since the length of the string and linear mass - density remain constant, and $\lambda$ is also constant for a given vibrating mode, $f\propto\sqrt{T}$.

Step2: Let the initial tension be $T_1$ and final tension be $T_2$

We know that $\frac{T_2}{T_1}=(1 - 0.036)=0.964$. Since $f\propto\sqrt{T}$, we have $\frac{f_2}{f_1}=\sqrt{\frac{T_2}{T_1}}$. Given $f_1 = 294$ Hz.

Step3: Calculate the new frequency $f_2$

$f_2=f_1\sqrt{\frac{T_2}{T_1}}=294\times\sqrt{0.964}$.
\[

$$\begin{align*} \sqrt{0.964}&\approx0.982\\ f_2&=294\times0.982 = 294\times(1 - 0.018)=294-294\times0.018\\ &=294 - 5.292\approx289\text{ Hz} \end{align*}$$

\]

Step4: Calculate the beat frequency $f_{beat}$

The beat frequency $f_{beat}=\vert f_1 - f_2\vert$. Since $f_1 = 294$ Hz and $f_2\approx289$ Hz, $f_{beat}=294 - 289=5.2$ Hz. Rounding to two significant figures, $f_{beat}=5.2$ Hz.

Answer:

$5.2$ Hz