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Question: A piano tuner has a tuning fork that sounds with a frequency of \(250\)Hz. The tuner strikes the for...

A piano tuner has a tuning fork that sounds with a frequency of 250250Hz. The tuner strikes the fork and plays a key that sounds with a frequency of 200200Hz. What is the frequency of the beats that the piano tuner hears?
(a) 0 Hz
(b) 0.8 Hz
(c) 1.25 Hz
(d) 50 Hz
(e) 450 Hz

Explanation

Solution

In this solution, we are going to use the difference in two frequencies to find the frequency of beats. Both the frequencies are given in question, we just need to find their difference.

Complete Step by Step Answer: Given:
Frequency of the tuning fork f1=250{f_1} = 250Hz
Frequency of the piano key played f2=200{f_2} = 200Hz
The frequency of beatsffthat the piano tuner hears is nothing but the difference of the frequencies of the two waves that are interfering to produce the Beats. In this question, the two waves are of tuning fork and piano key.
The formula to find the frequency of beatsffis given as:
The frequency of beatsf=f1f2      ...........(1)f = \left| {{f_1} - {f_2}} \right|\;\;\;...........(1)
Substituting the value off1=250{f_1} = 250Hz and f2=200{f_2} = 200Hz in equation (1), we get,
f=250  Hz200  Hz =50  Hz\begin{array}{c} f = \left| {250\;{\rm{Hz}} - 200\;{\rm{Hz}}} \right|\\\ = 50\;{\rm{Hz}} \end{array}
Therefore, the frequency of beats is 5050Hz. So option (d) is our correct option.
Additional Information: A piano tuner uses the interesting phenomenon of Beats to tune his piano using a tuning fork. The tuner produces the sound of tuning fork and piano at the same time, and the two sounds are heard distinctively, we call it beats. The tuner then changes the tension of piano strings such that the beats frequency becomes zero, and one common sound must be heard when the tuning fork and the piano are played together.

Note: The modulus (two vertical bars) in the formula of frequency of the beats represent that the difference of the two frequencies of the interfering notes will always be positive (as frequency can never be negative). So, be aware that you can calculate the difference of frequencies as (f1f2)\left( {{f_1} - {f_2}} \right) or(f2f1)\left( {{f_2} - {f_1}} \right) but the resulting frequency of the beats must always be given with a positive sign.