(a) 4A
In a simple harmonic motion:
- The amplitude (A) is the maximum displacement of the system, measured with respect to the equilibrium position
- The period (T) is the time needed for one complete oscillation, so for instance is the time the system needs to go from position x=+A back to x=+A again
Therefore, we have that in one time period (1T) the distance covered is 4A. In fact, during one period (1T), the system:
- Goes from x=+A to x=0 (equilibrium position) --> distance covered: A
- Goes from x=0 to x=-A --> distance covered: A
- Goes from x=-A to x=0 (equilibrium position) --> distance covered: A
- Goes from x=0 to x=+A --> distance covered: A
So, in total, 4A.
(b) 20A
Since the system moves through a distance of 4A in a time interval of 1T, we can set a proportion to see what is the distance covered in the time 5.00 T:
![1 T : 4 A = 5T : d](https://tex.z-dn.net/?f=1%20T%20%3A%204%20A%20%3D%205T%20%3A%20d)
Solving for d, we find
![d=\frac{(4A)(5T)}{1 T}=20A](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B%284A%29%285T%29%7D%7B1%20T%7D%3D20A)
So, the distance covered in the time 5.00 T is 20 A.
(c) 0.5 T
Since the system moves through a distance of 4A in a time interval of 1T, we can set a proportion to see the time t that the system needs to move through a total distance of 2A:
![1 T : 4 A = t : 2A](https://tex.z-dn.net/?f=1%20T%20%3A%204%20A%20%3D%20t%20%3A%202A)
Solving for t, we find
![t=\frac{(2A)(1T)}{4 A}=0.5 T](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%282A%29%281T%29%7D%7B4%20A%7D%3D0.5%20T)
So, the time needed for the system to move through a total distance of 2A is 0.5T (half period).
(d) 7/4 T
As before, since the system moves through a distance of 4A in a time interval of 1T, we can set a proportion to see the time t that the system needs to move through a total distance of 7A:
![1 T : 4 A = t : 7A](https://tex.z-dn.net/?f=1%20T%20%3A%204%20A%20%3D%20t%20%3A%207A)
Solving for t, we find
![t=\frac{(7A)(1T)}{4 A}=\frac{7}{4}T](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%287A%29%281T%29%7D%7B4%20A%7D%3D%5Cfrac%7B7%7D%7B4%7DT)
So, the time needed for the system to move through a total distance of 2A is 7/4 T
(e) 8/5 D
In a time of
, the distance covered is 16D.
We also now that the distance covered in 1T is 4A.
So we can find the distance covered in a time of
in terms of A:
![1T:4A = \frac{5}{2}T:d\\d=\frac{(4A)(\frac{5}{2}T)}{1T}=10A](https://tex.z-dn.net/?f=1T%3A4A%20%3D%20%5Cfrac%7B5%7D%7B2%7DT%3Ad%5C%5Cd%3D%5Cfrac%7B%284A%29%28%5Cfrac%7B5%7D%7B2%7DT%29%7D%7B1T%7D%3D10A)
And we know that this distance must correspond to 16D, so we can find a relationship between A and D:
![10A=16D\\A=\frac{16}{10}D=\frac{8}{5}D](https://tex.z-dn.net/?f=10A%3D16D%5C%5CA%3D%5Cfrac%7B16%7D%7B10%7DD%3D%5Cfrac%7B8%7D%7B5%7DD)