(a) first two overtones for each string:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
![f_n = n f_1](https://tex.z-dn.net/?f=f_n%20%3D%20n%20f_1)
where f1 is the fundamental frequency.
So, the first overtone (2nd harmonic) of the string is
![f_2 = 2 f_1 = 2 \cdot 196 Hz = 392 Hz](https://tex.z-dn.net/?f=f_2%20%3D%202%20f_1%20%3D%202%20%5Ccdot%20196%20Hz%20%3D%20392%20Hz)
while the second overtone (3rd harmonic) is
![f_3 = 3 f_1 = 3 \cdot 196 Hz = 588 Hz](https://tex.z-dn.net/?f=f_3%20%3D%203%20f_1%20%3D%203%20%5Ccdot%20196%20Hz%20%3D%20588%20Hz)
Similarly, for the second string with fundamental frequency
![f_1 = 523 Hz](https://tex.z-dn.net/?f=f_1%20%3D%20523%20Hz)
, the first overtone is
![f_2 = 2 f_1 = 2 \cdot 523 Hz = 1046 Hz](https://tex.z-dn.net/?f=f_2%20%3D%202%20f_1%20%3D%202%20%5Ccdot%20523%20Hz%20%3D%201046%20Hz)
and the second overtone is
![f_3 = 3 f_1 = 3 \cdot 523 Hz = 1569 Hz](https://tex.z-dn.net/?f=f_3%20%3D%203%20f_1%20%3D%203%20%5Ccdot%20523%20Hz%20%3D%201569%20Hz)
(b) The fundamental frequency of a string is given by
![f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }](https://tex.z-dn.net/?f=f%3D%20%20%5Cfrac%7B1%7D%7B2L%7D%20%20%5Csqrt%7B%20%5Cfrac%7BT%7D%7B%5Cmu%7D%20%7D%20%20)
where L is the string length, T the tension, and
![\mu = m/L](https://tex.z-dn.net/?f=%5Cmu%20%3D%20m%2FL)
is the mass per unit of length. This part of the problem says that the tension T and the length L of the string are the same, while the masses are different (let's calle them
![m_{196}](https://tex.z-dn.net/?f=m_%7B196%7D)
, the mass of the string of frequency 196 Hz, and
![m_{523}](https://tex.z-dn.net/?f=m_%7B523%7D)
, the mass of the string of frequency 523 Hz.
The ratio between the fundamental frequencies of the two strings is therefore:
![\frac{523 Hz}{196 Hz} = \frac{ \frac{1}{2L} \sqrt{ \frac{T}{m_{523}/L} } }{\frac{1}{2L} \sqrt{ \frac{T}{m_{196}/L} }}](https://tex.z-dn.net/?f=%20%5Cfrac%7B523%20Hz%7D%7B196%20Hz%7D%20%3D%20%20%5Cfrac%7B%20%5Cfrac%7B1%7D%7B2L%7D%20%20%5Csqrt%7B%20%5Cfrac%7BT%7D%7Bm_%7B523%7D%2FL%7D%20%7D%20%7D%7B%5Cfrac%7B1%7D%7B2L%7D%20%20%5Csqrt%7B%20%5Cfrac%7BT%7D%7Bm_%7B196%7D%2FL%7D%20%7D%7D%20%20)
and since L and T simplify in the equation, we can find the ratio between the two masses:
![\frac{m_{196}}{m_{523}}=( \frac{523 Hz}{196 Hz} )^2 = 7.1](https://tex.z-dn.net/?f=%20%5Cfrac%7Bm_%7B196%7D%7D%7Bm_%7B523%7D%7D%3D%28%20%5Cfrac%7B523%20Hz%7D%7B196%20Hz%7D%20%29%5E2%20%3D%207.1%20)
(c) Now the tension T and the mass per unit of length
![\mu](https://tex.z-dn.net/?f=%5Cmu)
is the same for the strings, while the lengths are different (let's call them
![L_{196}](https://tex.z-dn.net/?f=L_%7B196%7D)
and
![L_{523}](https://tex.z-dn.net/?f=L_%7B523%7D)
). Let's write again the ratio between the two fundamental frequencies
And since T and
![\mu](https://tex.z-dn.net/?f=%5Cmu)
simplify, we get the ratio between the two lengths:
![\frac{L_{196}}{L_{523}}= \frac{523 Hz}{196 Hz}=2.67](https://tex.z-dn.net/?f=%20%5Cfrac%7BL_%7B196%7D%7D%7BL_%7B523%7D%7D%3D%20%5Cfrac%7B523%20Hz%7D%7B196%20Hz%7D%3D2.67%20%20%20)
(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them
![T_{196}](https://tex.z-dn.net/?f=T_%7B196%7D)
and
![T_{523}](https://tex.z-dn.net/?f=T_%7B523%7D)
. Let's write again the ratio of the frequencies:
![\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L} \sqrt{ \frac{T_{523}}{m/L} } }{\frac{1}{2L} \sqrt{ \frac{T_{196}}{m/L} }}](https://tex.z-dn.net/?f=%5Cfrac%7B523%20Hz%7D%7B196%20Hz%7D%3D%20%5Cfrac%7B%20%5Cfrac%7B1%7D%7B2L%7D%20%5Csqrt%7B%20%5Cfrac%7BT_%7B523%7D%7D%7Bm%2FL%7D%20%7D%20%7D%7B%5Cfrac%7B1%7D%7B2L%7D%20%5Csqrt%7B%20%5Cfrac%7BT_%7B196%7D%7D%7Bm%2FL%7D%20%7D%7D)
Now m and L simplify, and we get the ratio between the two tensions: