Amass moves back and forth in simple harmonic motion with amplitude a and period t.
a. in terms of t, how long does it take for the mass to move through a total distance of 2a?
t/t=
b. how long does it take for the mass to move through a total distance of 5a?
t/t=

a. 0.5 T

- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position

- The period T is the time the system takes to complete one oscillation

During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.

So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

and solving for t we find

b. 1.25T

Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that

- the mass takes a time of 1 T to cover a distance of 4A

we can set the following proportion:

And by solving for t, we find

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.

and 1and 1