Question 14.11: SOUR NOTES GOAL Apply the beat frequency concept. PROBLEM A ...
SOUR NOTES
GOAL Apply the beat frequency concept.
PROBLEM A certain piano string is supposed to vibrate at a frequency of 4.40 \times 10^{2} \mathrm{~Hz}. To check its frequency, a tuning fork known to vibrate at a frequency of 4.40 \times 10^{2} \mathrm{~Hz} is sounded at the same time the piano key is struck, and a beat frequency of 4 beats per second is heard. (a) Find the two possible frequencies at which the string could be vibrating. (b) Suppose the piano tuner runs toward the piano, hold ing the vibrating tuning fork while his assistant plays the note, which is at 436 \mathrm{~Hz}. At his maximum speed, the piano tuner notices the beat frequency drops from 4 \mathrm{~Hz} to 2 \mathrm{~Hz} (without going through a beat frequency of zero). How fast is he moving? Use a sound speed of 343 \mathrm{~m} / \mathrm{s}. (c) While the piano tuner is running, what beat frequency is observed by the assistant? Note: Assume all numbers are accurate to two decimal places, necessary for this last calculation.
STRATEGY (a) The beat frequency is equal to the absolute value of the difference in frequency between the two sources of sound and occurs if the piano string is tuned either too high or too low. Solve Equation 14.20
f_b = | f_2- f_1| [14.20]
for these two possible frequencies. (b) Moving toward the piano raises the observed piano string frequency. Solve the Doppler shift for mula, Equation 14.12, for the speed of the observer. (c) The assistant observes a Doppler shift for the tuning fork. Apply Equation 14.12.
f_{O}=f_{S}\left(\begin{array}{l}v+v_{O} \\ \hline v-v_{S}\end{array}\right) [14.12]
(a) Find the two possible frequencies.
Case 1: f_{2}-f_{1} is already positive, so just drop the absolute-value signs:
\begin{aligned}f_{b} &=f_{2}-f_{1} \rightarrow 4 \mathrm{~Hz}=f_{2} -4.40 \times 10^{2} \mathrm{~Hz} \\f_{2} &=444 \mathrm{~Hz} \end{aligned}
Case 2: f_{2}-f_{1} is negative, so drop the absolute-value signs, but apply an overall negative sign:
\begin{aligned}f_{b} &=-\left(f_{2}-f_{1}\right)-\rightarrow 4 \mathrm{~Hz}=-\left(f_{2}-4.40 \times 10^{2} \mathrm{~Hz}\right) \\f_{2} &=436 \mathrm{~Hz}\end{aligned}
(b) Find the speed of the observer if running toward the piano results in a beat frequency of 2 \mathrm{~Hz}.
Apply the Doppler shift to the case where frequency of the piano string heard by the running observer is f_{O}=438 \mathrm{~Hz}:
\begin{aligned}& f_{O}=f_{s}\left(\begin{array}{cc}v+v_{O} \\ \hline v – v_{S}\end{array}\right) \\& 438 \mathrm{~Hz}=(436 \mathrm{~Hz})\left(\frac{343 \mathrm{~m} / \mathrm{s}+v_{O}}{343 \mathrm{~m} / \mathrm{s}}\right) \\& v_{O}=\left(\frac{438 \mathrm{~Hz}-436 \mathrm{~Hz}}{436 \mathrm{~Hz}}\right)(343 \mathrm{~m} / \mathrm{s})=1.57 \mathrm{~m} / \mathrm{s} \end{aligned}
(c) What beat frequency does the assistant observe?
Apply Equation 14.12. Now the source is the tuning fork, so f_{S}=4.40 \times 10^{2} \mathrm{~Hz}.
\begin{aligned}\\& f_{O}=f_{S}\left(\begin{array}{l}v+v_{O} \\ \hline v-v_{S}\end{array}\right) \\& =\left(440 \times 10^{2} \mathrm{~Hz}\right)\left(\begin{array}{c}343 \mathrm{~m} / \mathrm{s} \\ \hline 343 \mathrm{~m} / \mathrm{s}-1.57 \mathrm{~m} / \mathrm{s}\end{array}\right)=442 \mathrm{~Hz} \end{aligned}
Compute the beat frequency:
f_{b}=f_{2}-f_{1}=442 \mathrm{~Hz}-436 \mathrm{~Hz}=6 \mathrm{~Hz}
REMARKS The assistant on the piano bench and the tuner running with the fork observe different beat frequencies. Many physical observations depend on the state of motion of the observer, a subject discussed more fully in Topic 26, on relativity.