(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:
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:
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:
so, it takes 1.75 T.
(e)
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
The problem also says us that distance is equal to
d = 16 D
So by combining the two equations, we find