A violinist is tuning the A string on her violin by listening for beats when this note is played simultaneously with a tuning fork of frequency 440 Hz.She hears a beat frequency of 4 Hz.She notices that,when she increases the tension in the string slightly,the beat frequency decreases.What was the frequency of the mistuned A string?
A)448 Hz
B)444 Hz
C)436 Hz
D)432 Hz
E)438 Hz

Question

Quizplus · Accepted Answer

When the violinist and the tuning fork are played together, they create a phenomenon called beats where two sounds of slightly different frequencies interfere with each other, creating a periodic variation in loudness. Let's say the violinist originally played a note with a frequency of x Hz. When played with the tuning fork of 440 Hz, the beat frequency is given by the absolute value of the difference between the two frequencies: |440 - x| Hz.

In this case, the beat frequency is given to be 4 Hz. So, we have:

|440 - x| = 4

Now, if the violinist increases the tension in the string slightly, the frequency of the note will increase. Let's say the new frequency is y Hz. Since we know the beat frequency is decreasing, we now have:

|440 - y| < 4

Subtracting the first equation from the second, we get:

|y - x| < 4

Since we know that xy.

Therefore, the frequency of the mistuned A string is closest to 436 Hz, which is option C.

A Violinist Is Tuning the a String on Her Violin