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A weight attached to a spring is at its lowest point, 9 inches below equilibrium, at time [tex]\( t=0 \)[/tex] seconds. When the weight is released, it oscillates and returns to its original position at [tex]\( t=3 \)[/tex] seconds.

Which of the following equations models the distance, [tex]\( d \)[/tex], of the weight from its equilibrium after [tex]\( t \)[/tex] seconds?

A. [tex]\( d = -9 \cos \left(\frac{\pi}{3} t\right) \)[/tex]

B. [tex]\( d = -9 \cos \left(\frac{2 \pi}{3} t\right) \)[/tex]

C. [tex]\( d = -3 \cos \left(\frac{\pi}{9} t\right) \)[/tex]

D. [tex]\( d = -3 \cos \left(\frac{2 \pi}{9} t\right) \)[/tex]

Sagot :

GinnyAnsweridn
To determine which equation models the distance [tex]\( d \)[/tex] of the weight from its equilibrium after [tex]\( t \)[/tex] seconds, we need to find an equation that satisfies the conditions provided: the weight is 9 inches below equilibrium at [tex]\( t = 0 \)[/tex] and returns to this position after [tex]\( t = 3 \)[/tex] seconds.

Given equations:
1. [tex]\( d = -9 \cos \left( \frac{\pi}{3} t \right) \)[/tex]
2. [tex]\( d = -9 \cos \left( \frac{2\pi}{3} t \right) \)[/tex]
3. [tex]\( d = -3 \cos \left( \frac{\pi}{9} t \right) \)[/tex]
4. [tex]\( d = -3 \cos \left( \frac{2\pi}{9} t \right) \)[/tex]

Let's test each equation separately:

### 1. [tex]\( d = -9 \cos \left( \frac{\pi}{3} t \right) \)[/tex]

At [tex]\( t = 0 \)[/tex]:
[tex]\[ d = -9 \cos \left( \frac{\pi}{3} \times 0 \right) = -9 \cos(0) = -9 \times 1 = -9 \][/tex]
This satisfies the condition at [tex]\( t = 0 \)[/tex].

At [tex]\( t = 3 \)[/tex]:
[tex]\[ d = -9 \cos \left( \frac{\pi}{3} \times 3 \right) = -9 \cos(\pi) = -9 \times -1 = 9 \][/tex]
This does not match the condition at [tex]\( t = 3 \)[/tex] since we need [tex]\( d = -9 \)[/tex].

### 2. [tex]\( d = -9 \cos \left( \frac{2\pi}{3} t \right) \)[/tex]

At [tex]\( t = 0 \)[/tex]:
[tex]\[ d = -9 \cos \left( \frac{2\pi}{3} \times 0 \right) = -9 \cos(0) = -9 \times 1 = -9 \][/tex]
This satisfies the condition at [tex]\( t = 0 \)[/tex].

At [tex]\( t = 3 \)[/tex]:
[tex]\[ d = -9 \cos \left( \frac{2\pi}{3} \times 3 \right) = -9 \cos(2\pi) = -9 \times 1 = -9 \][/tex]
This satisfies the condition at [tex]\( t = 3 \)[/tex].

### 3. [tex]\( d = -3 \cos \left( \frac{\pi}{9} t \right) \)[/tex]

At [tex]\( t = 0 \)[/tex]:
[tex]\[ d = -3 \cos \left( \frac{\pi}{9} \times 0 \right) = -3 \cos(0) = -3 \times 1 = -3 \][/tex]
This does not match the condition at [tex]\( t = 0 \)[/tex].

### 4. [tex]\( d = -3 \cos \left( \frac{2\pi}{9} t \right) \)[/tex]

At [tex]\( t = 0 \)[/tex]:
[tex]\[ d = -3 \cos \left( \frac{2\pi}{9} \times 0 \right) = -3 \cos(0) = -3 \times 1 = -3 \][/tex]
This does not match the condition at [tex]\( t = 0 \)[/tex].

Only the second equation [tex]\( d = -9 \cos \left( \frac{2\pi}{3} t \right) \)[/tex] satisfies the condition both at [tex]\( t = 0 \)[/tex] and [tex]\( t = 3 \)[/tex] seconds. Thus, the correct equation that models the distance [tex]\( d \)[/tex] of the weight from its equilibrium after [tex]\( t \)[/tex] seconds is:

[tex]\[ d = -9 \cos \left( \frac{2\pi}{3} t \right) \][/tex]
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