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The mass of the first block will be three times the mass of the second block.
According to Newton's second law of motion, the force acting on a body is directly proportional to the acceleration as shown;
F is the acting force
m is the mass
a is the acceleration of the body
Given the following parameters
Constant force F = 1N
For the first block with the acceleration of "a"
1 = m₁a
a = m₁/1
m₁ = a .................1
For the second block, acceleration is thrice that of the first. This means;
F = m(3a)
1 = 3ma
..........................2
Divide both equations
From the calculation, we can conclude that the mass of the first block will be three times the mass of the second block.
Learn more here: brainly.com/question/19030143
(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: