Question

watch the video and then solve the problem given below. click here to watch the video. an object is attached to a coiled spring. the object is pulled down (negative direction from the rest position) 5 centimeters, and then released. write an equation for the distance d of the object from its rest position, after t seconds if the amplitude is 5 centimeters and the period is 5 seconds. the equation for the distance d of the object from its rest position is (type an exact answer, using π as needed. use integers or fractions for any numbers in the equation.)

Explanation:

Step1: Recall the general form of a sinusoidal - type equation for simple harmonic motion

The general form of the equation for the displacement $d$ of an object in simple - harmonic motion is $d = A\sin(\omega t+\varphi)$ or $d = A\cos(\omega t+\varphi)$, where $A$ is the amplitude, $\omega$ is the angular frequency, $t$ is the time, and $\varphi$ is the phase shift. Since the object is pulled down (negative direction) from the rest position and released, we use the cosine function $d = A\cos(\omega t+\varphi)$. At $t = 0$, $d=-A$. When $t = 0$, $d = A\cos(\varphi)$. Substituting $d=-A$ and $t = 0$ into $d = A\cos(\omega t+\varphi)$, we get $-A=A\cos(\varphi)$, so $\cos(\varphi)= - 1$ and $\varphi=\pi$.

Step2: Calculate the angular frequency $\omega$

The formula for the period $T$ of a sinusoidal function is $T=\frac{2\pi}{\omega}$. Given that $T = 5$ seconds, we can solve for $\omega$. Rearranging the formula $T=\frac{2\pi}{\omega}$ for $\omega$, we get $\omega=\frac{2\pi}{T}$. Substituting $T = 5$ into the formula, we have $\omega=\frac{2\pi}{5}$.

Step3: Write the equation

The amplitude $A = 5$ centimeters. Substituting $A = 5$, $\omega=\frac{2\pi}{5}$, and $\varphi=\pi$ into the equation $d = A\cos(\omega t+\varphi)$, we get $d = 5\cos(\frac{2\pi}{5}t+\pi)$. Using the trigonometric identity $\cos(a + \pi)=-\cos(a)$, we can rewrite the equation as $d=-5\cos(\frac{2\pi}{5}t)$.

Answer:

$d=-5\cos(\frac{2\pi}{5}t)$