Two violists are tuning their A strings, for which the fundamental frequency is

Question

Two violists are tuning their A strings, for which the fundamental frequency is 440 Hz (properly tuned). 3 beats per second is heard when they both play the A note.

a) By what fractoinal amount must the player of the viola with the lower frequency increase the tension in their string to make its frequency identical to that of the other viola? Assume that the other viola is already properly tuned with the A string under the tension of 520 N.

b) If instead of increasing the tension in the string, the second violist decreases the tension by the same amount, what will be the frequency of the beats?

Explanation / Answer

consider linear mass density is mu and wavelength will also be same.

speed of wave = sqrt(T / mu)

and frquency = speed / wavelength

frequency is direclty propotional to sqrt(T).

beat = frequency difference

Tension is increases so frequency of untuned will be 440 +3 = 443 Hz

hence, f1/f2 = T1/T2

443/440 = sqrt(T / 520 )

T = 527.12 N

b) Tension difference = 527.12 - 520 = 7.12 N

new Tension, T = 520 - 7.12 = 512.88 N

now using f1/f2 = T1/T2

f1 / 440 = 512.88/520

f1 = 434 Hz

beat = 440 - 434 = 6 beats per second.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.