Physics, 30.11.2019 08:31 Manuel2019

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? ? a(b) through what distance does it move in the time 5.00t? ? a(c) in terms of t, how long does it take for the mass to move through a total distance of 2a? ? t(d) how long does it take for the mass to move through a total distance of 7a? ? t(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? ? d

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Answer from: yami85

(a) 4A

In a simple harmonic motion:

- The amplitude (A) is the maximum displacement of the system, measured with respect to the equilibrium position

- The period (T) is the time needed for one complete oscillation, so for instance is the time the system needs to go from position x=+A back to x=+A again

Therefore, we have that in one time period (1T) the distance covered is 4A. In fact, during one period (1T), the system:

- Goes from x=+A to x=0 (equilibrium position) --> distance covered: A

- Goes from x=0 to x=-A --> distance covered: A

- Goes from x=-A to x=0 (equilibrium position) --> distance covered: A

- Goes from x=0 to x=+A --> distance covered: A

So, in total, 4A.

(b) 20A

Since the system moves through a distance of 4A in a time interval of 1T, we can set a proportion to see what is the distance covered in the time 5.00 T:

1 T : 4 A = 5T : d

Solving for d, we find

$d=\frac{(4A)(5T)}{1 T}=20A$

So, the distance covered in the time 5.00 T is 20 A.

Since the system moves through a distance of 4A in a time interval of 1T, we can set a proportion to see the time t that the system needs to move through a total distance of 2A:

1 T : 4 A = t : 2A

Solving for t, we find

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

So, the time needed for the system to move through a total distance of 2A is 0.5T (half period).

(d) 7/4 T

As before, since the system moves through a distance of 4A in a time interval of 1T, we can set a proportion to see the time t that the system needs to move through a total distance of 7A:

1 T : 4 A = t : 7A

Solving for t, we find

$t=\frac{(7A)(1T)}{4 A}=\frac{7}{4}T$

So, the time needed for the system to move through a total distance of 2A is 7/4 T

(e) 8/5 D

In a time of $\frac{5}{2}T$ , the distance covered is 16D.

We also now that the distance covered in 1T is 4A.

So we can find the distance covered in a time of $\frac{5}{2}T$ in terms of A:

$1T:4A = \frac{5}{2}T:d\\d=\frac{(4A)(\frac{5}{2}T)}{1T}=10A$

And we know that this distance must correspond to 16D, so we can find a relationship between A and D:

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

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