Question

13.18 Two piano strings with the same mass and the same length are to be tuned to 440 Hz (A on the equal temperament scale). One string, with a tension of 300 N is already tuned to 440 Hz. The second string frequency is 435 Hz. a) Should the tension of string 2 be increased or decreased to tune the frequency to 440 Hz? b) How much must the tension of string 2 be changed to change the frequency from 435 Hz to 440 Hz?

14.2 The wavelength of a 450 Hz sound wave in air has a wavelength of 0.78 m. a) What is the speed of sound in air? b) What is the temperature?

Answer 1

Given

two pianon strings of the same mass and the same length are to be tuned to 440 Hz

the string 1 tuned to 300 N of the frequency is f1 = 440 Hz

string 2 with frequncy 435 Hz , to tune the string2 to 440 Hz we have to increased te tension

because the frequency of the string fixed at both ends is  

f1 = v/2l

and the speed of the wave on a string is v = sqrt(T/mue)

here mue is same so

f1 = (1/2l)(sqrt(T/mue))

that is frequency is proportional to the tension  

f2/f1 = sqrt(T2/T1)

a)

so to increase frequency from 435 Hz to 440 Hz , we should increase the tension  

b)

f2^2/f1^2 = T2/T1

440^2/435^2 = T2/300

T2 = 306.94 N

14.2

wavelength of sound wave in air is lambda = 0.78 m

frequency is f = 450 Hz ,

a)

the speed of sound in air is V = lambda*f = 0.78*450m/s = 351 m/s

b)

we know the formula is V = 331 m/s +0.61 m/s /C *T

T is the temperature

351 = 331+0.61*T

T = 32.786 0C