A weight is attached to a spring suspended vertically from a ceiling. When a dri

Question

A weight is attached to a spring suspended vertically from a ceiling. When a driving force is applied to the system, the weight moves vertically from its equilibrium position, and this motion is modeled by y=1/3sin2t+1/4cos2t where y is the distance from equilibrium (in feet) and t is the time (in seconds). Use the identity a*sinBtheta+b*cosBtheta=square root a^2+b^2*sin(Bt+C)

1. Find the amplitude of the oscillations of the weight

2. Find the frequency of the oscillations of the weight.

3. I really want to understand this.

Explanation / Answer

Using the identity, match up the numbers to the identity:

1/3    sin2t       +   1/4    cos2t

a  * sinBtheta +    b * cosBtheta

So:

a = 1/3

b = 1/4

B = 2

So, for the second part of the identity:

a*sinBtheta+b*cosBtheta=square root a^2+b^2*sin(Bt+C)

Plug the numbers in:

`sqrt((1/3)^(2)+(1/4)^(2))` *sin(2t+C) = (25/144) * sin(2t+C)

So, the amplitude is the number in front of the trig function,

amplitude = 25/144

And, the frequency is the number in front of the "t" in the argument of the trig function,

frequency = 2

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