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The motion of a simple spring hanging from the ceiling can be modeled with a cosine function. The bottom of the spring has a maximum height of 7 feet 4 inches and a minimum height of 6 feet 2 inches from the floor. It takes 2 seconds for the spring to expand from its minimum length to its maximum length. What is a cosine function that models the spring’s length in inches above and below its average, resting position? Express the model as a function of time in seconds

Verified Answer :

meerkat18

The first step is to substitute the given values in this equation f(x)= A cos (W*t). It is assumed that there is no mass in the resting position. The calculated amplitude is equal to 7. The final answer is f(t) = 7cos(π/2t).

Answer:

\(y = 6'9" + 7"cos(\frac{\pi}{2} t)\)

Explanation:

As we know that the distance between maximum and minimum distance of the position will be equal to double of the amplitude.

So here we can say that

\(2 \times Amplitude = maximum\: distance - minimum \:distance\)

\(2A = 7'4" - 6'2"\)

\(2A = 1'2"\)

\(A = 7 inch\)

Since it took 2 second to reach the position of maximum length from its position of minimum length so here time period of motion will be

\(T = 4 seconds\)

so here angular frequency is given as

\(\omega = \frac{2\pi}{T}\)

\(\omega = \frac{2\pi}{4}\)

now the equation of motion will be

\(y = 6'9" + 7"cos(\frac{\pi}{2} t)\)

here its mean position from ground is at 6 ft 9 inch above and it will oscillate about it with an amplitude of 7 inch with time period of 4 s

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