Question

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

Answer 1

(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:

$1.0 T : 4 A = 5.0 T : d\\d=\frac{(4A)(5.0T)}{1.0 T}=20 A$

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:

$1.0 T:4A=t : 2A\\t=\frac{(1.0T)(2A)}{4A}=0.5 T$

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:

$1.0 T:4A=t : 7A\\t=\frac{(1.0T)(7A)}{4A}=1.75 T$

so, it takes 1.75 T.

(e) $\frac{8}{5}D$

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

$1.0T:4 A=\frac{5}{2}T:d\\d=\frac{(4A)(\frac{5}{2}T)}{1.0 T}=10 A$

The problem also says us that distance is equal to

d = 16 D

So by combining the two equations, we find

$d=10 A=16 D\\A=\frac{16}{10}D=\frac{8}{5}D$