Question

post test: trigonometric functions
14
select the correct answer.
jackson is conducting an experiment for his physics class. he attaches a weight to the bottom of a metal spring. he then pulls the weight down so that it is a distance of 6 inches from its equilibrium position. jackson then releases the weight and finds that it takes 4 seconds for the spring to complete one oscillation.
which function best models the position of the weight?
a. s(t)=6sin(\frac{\pi}{2}t)
b. s(t)=6sin(2\pi t)
c. s(t)= - 6cos(2\pi t)
d. s(t)= - 6cos(\frac{\pi}{2}t)

Explanation:

Step1: Recall the general form of a harmonic - motion function

The general form of a harmonic - motion function for a spring - mass system is $s(t)=A\cos(\omega t)$ or $s(t)=A\sin(\omega t)$, where $A$ is the amplitude and $\omega$ is the angular frequency. The amplitude $A$ is the maximum displacement from the equilibrium position. Here, the weight is pulled down 6 inches from its equilibrium position, so $A = 6$.

Step2: Calculate the angular frequency $\omega$

The period $T$ is the time it takes to complete one oscillation. Given that $T = 4$ seconds. The formula for the angular frequency is $\omega=\frac{2\pi}{T}$. Substituting $T = 4$ into the formula, we get $\omega=\frac{2\pi}{4}=\frac{\pi}{2}$.

Step3: Determine the correct function

At $t = 0$, the weight is at its maximum displacement (pulled down), and for a cosine - function $y = A\cos(\omega t)$, when $t = 0$, $y=A$ (for $A>0$), and for a sine - function $y = A\sin(\omega t)$, when $t = 0$, $y = 0$. Since the weight starts at its maximum displacement (downward, so we can use a negative cosine function to represent the downward direction), the function is $s(t)=-A\cos(\omega t)$. Substituting $A = 6$ and $\omega=\frac{\pi}{2}$, we get $s(t)=-6\cos(\frac{\pi}{2}t)$.

Answer:

D. $s(t)=-6\cos(\frac{\pi}{2}t)$