Question

a weight is attached to a spring and reaches its equilibrium position (x = 0). it is then set in motion resulting in a displacement of x = 10 cos t, where x is measured in centimeters and t is measured in seconds. see the figure shown to the right. answer parts (a) and (b). (type an integer or decimal rounded to one decimal place as needed.) what is the springs velocity when t = $\frac{pi}{3}$? - 8.7 cm/sec (type an integer or decimal rounded to one decimal place as needed.) what is the springs velocity when t = $\frac{3pi}{4}$? cm/sec (type an integer or decimal rounded to one decimal place as needed.)

Explanation:

Step1: Recall velocity - displacement relation

Velocity $v$ is the derivative of displacement $x$ with respect to time $t$. Given $x = 10\cos t$, then $v=\frac{dx}{dt}=- 10\sin t$ (using the derivative formula $\frac{d}{dt}(\cos t)=-\sin t$).

Step2: Calculate velocity at $t = \frac{3\pi}{4}$

Substitute $t=\frac{3\pi}{4}$ into the velocity formula $v=-10\sin t$. So $v=-10\sin(\frac{3\pi}{4})$. Since $\sin(\frac{3\pi}{4})=\frac{\sqrt{2}}{2}$, then $v=-10\times\frac{\sqrt{2}}{2}=-5\sqrt{2}\approx - 7.1$.

Answer:

$-7.1$ cm/sec