Forces are expressed in ________. (newtons or mass)
2 answers:
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(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 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
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
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
And since T and
(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them
Now m and L simplify, and we get the ratio between the two tensions:
Answer:
a) F_net = 6.48 10⁻¹⁸ ( ), b) x = 0.15 m
Explanation:
a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.
The electric force is given by Coulomb's law
F =
Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.
F_net= ∑ F = F₁ + F₂
let's locate a reference system in the load that is on the left side, the distances are
left side - electron r₁ = x
right side -electron r₂ = d-x
let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q
we substitute
F_net = k q Q ( )
F _net = kqQ ( )
let's substitute the values
F_net = 9 10⁹ 1.6 10⁻¹⁹ 4.50 10⁻⁹ ( )
F_net = 6.48 10⁻¹⁸ ( )
now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values of the distance (x) and the net force are given
x (m) F (N)
0.05 27.0 10-16
0.10 8.10 10-16
0.15 5.76 10-16
0.20 8.10 10-16
0.25 27.0 10-16
b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m
