Question

A mass moves back and forth in simple harmonic motion with amplitude A and period T.

Answer 1

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

$1 T : 4 A = t : 2 A$

and solving for t we find

$t=\frac{(1T)(2 A)}{4A}=0.5 T$

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:

$1 T : 4 A = t : 5 A$

And by solving for t, we find

$t=\frac{(1T)(5 A)}{4A}=\frac{5}{4} T=1.25 T$