Amass moves back and forth in simple harmonic motion with amplitude a and period t.
(a) in terms of a, through what distance does the mass move in the time t? (b) through what distance does it move in the time 5.00t? (c) in terms of t, how long does it take for the mass to move through a total distance of 2a? (d) how long does it take for the mass to move through a total distance of 7a? (e) if the objects undergoes simple harmonic motion with a period t. in the time 5t/2 the object moves through a total distance of 16d. in terms of d, what is the object's amplitude of motion?

(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

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