Two strings on a musical instrument are tuned to play at 196 hz (g) and 523 hz (c). (a) what are the first two overtones for each string? first overtone for 196 hz (g) hz second overtone for 196 hz (g) hz first overtone for 523 hz (c) hz second overtone for 523 hz (c) hz (b) if the two strings have the same length and are under the same tension, what must be the ratio of their masses (m196/m523)? (c) if the strings, instead, have the same mass per unit length and are under the same tension, what is the ratio of their lengths (l196/l523)? 1.33 2.67 (d) if their masses and lengths are the same, what must be the ratio of the tensions in the two strings? (t196/t523) 45 0.14

(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

where f1 is the fundamental frequency.

So, the first overtone (2nd harmonic) of the string is

while the second overtone (3rd harmonic) is


Similarly, for the second string with fundamental frequency , the first overtone is

and the second overtone is


(b) The fundamental frequency of a string is given by

where L is the string length, T the tension, and  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 , the mass of the string of frequency 196 Hz, and , the mass of the string of frequency 523 Hz.
The ratio between the fundamental frequencies of the two strings is therefore:

and since L and T simplify in the equation, we can find the ratio between the two masses:


(c) Now the tension T and the mass per unit of length  is the same for the strings, while the lengths are different (let's call them  and ). Let's write again the ratio between the two fundamental frequencies
 
And since T and  simplify, we get the ratio between the two lengths:


(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them  and . Let's write again the ratio of the frequencies:

Now m and L simplify, and we get the ratio between the two tensions:

The car would move to the right with 150 n of force. the 150 n of force from friction and air resistance slows the car from moving to the right with 300 n of force.

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