An object is attached by a string to the end of a spring. Fang throws the object upwards and starts a stopwatch at $t=0$ seconds. The object starts oscillating vertically in a periodic way that can be modeled by a trigonometric function.

The object's average height is $-20 \mathrm{~cm}$ (measured from the top of the spring). It first achieves that average height on the way up at $t=0.2$ seconds, and then again every 2 seconds. The object's maximum and minimum heights are each $5 \mathrm{~cm}$ from its average height.

Find the formula of the trigonometric function that models the height $H$ of the weight $t$ seconds after Fang started the stopwatch. Define the function using radians.
$$
H(t)=
$$
$\square$
What is the height of the object after 0.6 seconds? Round your answer, if necessary, to two decimal places.
$\square$ $\mathrm{cm}$

Question

An object is attached by a string to the end of a spring. Fang throws the object upwards and starts a stopwatch at $t=0$ seconds. The object starts oscillating vertically in a periodic way that can be modeled by a trigonometric function.

The object's average height is $-20 \mathrm{~cm}$ (measured from the top of the spring). It first achieves that average height on the way up at $t=0.2$ seconds, and then again every 2 seconds. The object's maximum and minimum heights are each $5 \mathrm{~cm}$ from its average height.

Find the formula of the trigonometric function that models the height $H$ of the weight $t$ seconds after Fang started the stopwatch. Define the function using radians.
$$
H(t)=
$$
$\square$
What is the height of the object after 0.6 seconds? Round your answer, if necessary, to two decimal places.
$\square$ $\mathrm{cm}$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the trigonometric function that models the height $H$ of the weight over time, we need to determine the amplitude, period, phase shift, and vertical shift. ‖ ‖ step 2 ⋮ The amplitude is the maximum deviation from the average height, which is $5 \mathrm{~cm}$. ‖ ‖ step 3 ⋮ The period is the time it takes for one complete oscillation, which is $2$ seconds. Since the function will be in terms of radians, we use the relationship $T = \frac{2\pi}{\omega}$ to find the angular frequency $\omega$. ‖ ‖ step 4 ⋮ Solving for $\omega$, we get $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi$. ‖ ‖ step 5 ⋮ The phase shift is determined by the time it first reaches the average height on the way up. Since this occurs at $t = 0.2$ seconds, the phase shift is $0.2$ seconds to the right. In terms of $\omega$, the phase shift is $\frac{\omega}{5} = \frac{\pi}{5}$. ‖ ‖ step 6 ⋮ The vertical shift is the average height, which is $-20 \mathrm{~cm}$. ‖ ‖ step 7 ⋮ Combining these values, the trigonometric function is $H(t) = 5 \cos(\pi t - \frac{\pi}{5}) - 20$. ‖ ‖ step 8 ⋮ To find the height of the object after $0.6$ seconds, we substitute $t = 0.6$ into the function. ‖ ‖ step 9 ⋮ Calculating $H(0.6) = 5 \cos(\pi \cdot 0.6 - \frac{\pi}{5}) - 20$. ‖ ‖ step 10 ⋮ Simplifying the cosine term, we get $H(0.6) = 5 \cos(\frac{3\pi}{5} - \frac{\pi}{5}) - 20 = 5 \cos(\frac{2\pi}{5}) - 20$. ‖ ‖ step 11 ⋮ Using a calculator, we find that $\cos(\frac{2\pi}{5}) \approx 0.309$. ‖ ‖ step 12 ⋮ Therefore, $H(0.6) \approx 5 \cdot 0.309 - 20 \approx -18.55 \mathrm{~cm}$. ‖ ∻[1] Answer∻ ⚹ $H(t) = 5 \cos(\pi t - \frac{\pi}{5}) - 20$ ⚹ ⚹ $H(0.6) \approx -18.55 \mathrm{~cm}$ ⚹ ∻Key Concept∻ ⚹Modeling periodic motion with trigonometric functions⚹ ∻Explanation∻ ⚹The trigonometric function for the height of the weight over time is determined by the amplitude, period, phase shift, and vertical shift. The height at a specific time is found by substituting the time into the function.⚹◊What is the formula for the trigonometric function that models the height of the weight over time?⍭ Generate me a similar question◊

Sia