(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:
![1.0 T : 4 A = 5.0 T : d\\d=\frac{(4A)(5.0T)}{1.0 T}=20 A](https://tex.z-dn.net/?f=1.0%20T%20%3A%204%20A%20%3D%205.0%20T%20%3A%20d%5C%5Cd%3D%5Cfrac%7B%284A%29%285.0T%29%7D%7B1.0%20T%7D%3D20%20A)
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:
![1.0 T:4A=t : 2A\\t=\frac{(1.0T)(2A)}{4A}=0.5 T](https://tex.z-dn.net/?f=1.0%20T%3A4A%3Dt%20%3A%202A%5C%5Ct%3D%5Cfrac%7B%281.0T%29%282A%29%7D%7B4A%7D%3D0.5%20T)
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:
![1.0 T:4A=t : 7A\\t=\frac{(1.0T)(7A)}{4A}=1.75 T](https://tex.z-dn.net/?f=1.0%20T%3A4A%3Dt%20%3A%207A%5C%5Ct%3D%5Cfrac%7B%281.0T%29%287A%29%7D%7B4A%7D%3D1.75%20T)
so, it takes 1.75 T.
(e) ![\frac{8}{5}D](https://tex.z-dn.net/?f=%5Cfrac%7B8%7D%7B5%7DD)
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
![1.0T:4 A=\frac{5}{2}T:d\\d=\frac{(4A)(\frac{5}{2}T)}{1.0 T}=10 A](https://tex.z-dn.net/?f=1.0T%3A4%20A%3D%5Cfrac%7B5%7D%7B2%7DT%3Ad%5C%5Cd%3D%5Cfrac%7B%284A%29%28%5Cfrac%7B5%7D%7B2%7DT%29%7D%7B1.0%20T%7D%3D10%20A)
The problem also says us that distance is equal to
d = 16 D
So by combining the two equations, we find
![d=10 A=16 D\\A=\frac{16}{10}D=\frac{8}{5}D](https://tex.z-dn.net/?f=d%3D10%20A%3D16%20D%5C%5CA%3D%5Cfrac%7B16%7D%7B10%7DD%3D%5Cfrac%7B8%7D%7B5%7DD)