An object is attached by a string to the end of a spring. Fang throws the object upwards and starts a stopwatch at t=0t=0t, equals, 0 seconds. The object starts oscillating vertically in a periodic way that can be modeled by a trigonometric function.
The object's average height is -20\text{ cm}−20 cmminus, 20, start text, space, c, m, end text (measured from the top of the spring). It first achieves that average height on the way up at t=0.2t=0.2t, equals, 0, point, 2 seconds, and then again every 222 seconds. The object's maximum and minimum heights are each 5\text{ cm}5 cm5, start text, space, c, m, end text from its average height.
Find the formula of the trigonometric function that models the height HHH of the weight ttt seconds after Fang started the stopwatch. Define the function using radians.
\qquad H(t) =H(t)=H, left parenthesis, t, right parenthesis, equals
What is the height of the object after 0.60.60, point, 6 seconds? Round your answer, if necessary, to two decimal places.

Respuesta :

The object attached to a string illustrates a sinusoidal trigonometric function

The trigonometric function is H(t) = 5sin(π/2(t-0.2))-20, and the object's height after 0.6 seconds is  -17.06 cm

How to determine the trigonometry function model?

The trigonometry model of the object is a sine function represented by:

H(t) = Asin(B(t - c)) + D

The maximum height is 5 cm.

So, we have:

Amplitude, A = 5

The average height is - 20 cm.

So, we have

Vertical shift, D = -20

Its first average height is at t = 0.2.

So, we have:

Horizontal shift, C = 0.2

The period B is then calculated as:

B = π/t

The object reaches the maximum height every 2 seconds.

So, we have:

B = π/2

Substitute these values in H(t) = Asin(B(t - c)) + D

H(t) = 5sin(π/2(t-0.2))-20

Hence, the trigonometric function is H(t) = 5sin(π/2(t-0.2))-20

The height of the object after 0.6 seconds

Substitute 0.6 for t in H(t)

H(0.6) = 5sin(π/2(0.6-0.2))-20

Evaluate

H(0.6) = -17.06

Hence, the object's height after 0.6 seconds is  -17.06 cm

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