post test: trigonometric functions\n14\nselect the correct answer.\njackson is conducting an experiment for…
post test: trigonometric functions\n14\nselect the correct answer.\njackson 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.\nwhich function best models the position of the weight?\na. s(t)=6sin(\\frac{\\pi}{2}t)\nb. s(t)=6sin(2\\pi t)\nc. s(t)= - 6cos(2\\pi t)\nd. s(t)= - 6cos(\\frac{\\pi}{2}t)
Answer
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. $s(t)=-6\cos(\frac{\pi}{2}t)$