Respuesta :
(a) 4A
In a simple harmonic motion:
- The amplitude is the maximum displacement of the system from the equilibrium position
- The period is the time that the system takes to make one complete oscillation: for example, the time it takes to go from a displacement of x=+A to the next x=+A.
So we have that the total distance covered by the system in one period T is 4 times the amplitude: 4A, because in one cycle the system does the following:
- Moves from x=+A to x=0 (equilibrium position) --> distance covered: A
- Moves from x=0 to x=-A --> distance covered so far: A+A=2A
- Moves from x=-A to x=0 (equilibrium position) --> distance covered so far: 2A+A=3A
- Moves from x=0 to x=+a --> distance covered so far: 3A+A=4A
(b) 20 A
We said that in a time of 1.0 T the system moves through a distance of 4A. Therefore, to find the distance d through which the system moves during a time of 5.0 T, we have to solve the following proportion:
[tex]1.0 T : 4 A = 5.0 T : d\\d=\frac{(4A)(5.0T)}{1.0 T}=20 A[/tex]
So, it moves through a distance of 20 A.
(c) 0.5 T
We said that in a time of 1.0 T the system moves through a distance of 4A. Therefore, to find the time t that it takes for the mass to move a total distance of 2A, we have to solve the following proportion:
[tex]1.0 T:4A=t : 2A\\t=\frac{(1.0T)(2A)}{4A}=0.5 T[/tex]
so, it takes half period.
(d) 1.75 T
We said that in a time of 1.0 T the system moves through a distance of 4A. Therefore, to find the time t that it takes for the mass to move a total distance of 7A, we have to solve the following proportion:
[tex]1.0 T:4A=t : 7A\\t=\frac{(1.0T)(7A)}{4A}=1.75 T[/tex]
so, it takes 1.75 T.
(e) [tex]\frac{8}{5}D[/tex]
We said that in a time of 1.0 T the system moves through a distance of 4A. So, the distance covered in a time of 5T/2 is given by the following proportion
[tex]1.0T:4 A=\frac{5}{2}T:d\\d=\frac{(4A)(\frac{5}{2}T)}{1.0 T}=10 A[/tex]
The problem also says us that distance is equal to
d = 16 D
So by combining the two equations, we find
[tex]d=10 A=16 D\\A=\frac{16}{10}D=\frac{8}{5}D[/tex]
a) The mass is at a distance of 0 when t = T.
b) The mass is at a distance of 0 when t = 5 · T.
c) A total distance of 2 · A is reached when t = 0.5 · T.
d) A total distance of 7 · A is reached when t = 1.75 · T.
e) The amplitude of motion of the object is 1.6 · D.
How to analyze a system is a simple harmonic motion
In this question we must apply the concepts of amplitude and period regarding the situation of a simple harmonic motion. A simple harmonic motion is a case of periodic motion in which effects from friction and air viscosity are negligible.
The amplitude is the magnitude of the maximum distance reached by the mass with respect to the equilibrium position and the period is the time need by the system under simple harmonic motion to complete one cycle.
There is the following relationships for the case of simple harmonic motion:
t = 0 → x = 0, t = 0.25 · T → x = + A, t = 0.5 · T → x = 0, t = 0.75 · T → x = -A, t = T → x = 0
We notice that the mass coves a distance of A each 0.25 · T. Now we proceed to answer each question:
a) Based on previous information, we conclude that the mass is at a distance of 0 when t = T. Â [tex]\blacksquare[/tex]
b) Based on previous information, we conclude that the mass is at a distance of 0 when t = 5 · T. [tex]\blacksquare[/tex]
c) By rule of three we have that a total distance of 2 · A is reached when t = 0.5 · T. [tex]\blacksquare[/tex]
d) By rule of three we have that a total distance of 7 · A is reached when t = 1.75 · T. [tex]\blacksquare[/tex]
e) Based on the fact that amplitude is reached from 0 to 0.25 · T, we determine the amplitude in terms of D by rule of three:
[tex]A = \frac{\frac{1}{4}\cdot T }{\frac{5}{2}\cdot T }\times 16\cdot D[/tex]
A = 1.6 · D
The amplitude of motion of the object is 1.6 · D. [tex]\blacksquare[/tex]
To learn more on simple harmonic motion, we kindly invite to check this verified question: https://brainly.com/question/17315536
