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). what is the springs displacement when t = 3π/4? (type an integer or decimal rounded to one decimal place as needed.) - 7.1 cm (type an integer or decimal rounded to one decimal place as needed.) (b) what is the springs velocity when t = 0? cm/sec (type an integer or decimal rounded to one decimal place as needed.)

Explanation:

Step1: Recall the displacement formula

The displacement is given by $x = 10\cos t$.

Step2: Find displacement at $t=\frac{3\pi}{4}$

Substitute $t = \frac{3\pi}{4}$ into $x = 10\cos t$. We know that $\cos(\frac{3\pi}{4})=-\frac{\sqrt{2}}{2}$. So $x = 10\times(-\frac{\sqrt{2}}{2})=- 5\sqrt{2}\approx - 7.1$ cm.

Step3: Recall the velocity - displacement relationship

Velocity $v$ is the derivative of displacement $x$ with respect to time $t$. The derivative of $x(t)=10\cos t$ using the rule $\frac{d}{dt}(\cos t)=-\sin t$ is $v(t)=x^\prime(t)=- 10\sin t$.

Step4: Find velocity at $t = 0$

Substitute $t = 0$ into $v(t)=-10\sin t$. Since $\sin(0) = 0$, then $v(0)=-10\times0 = 0$ cm/sec.

Answer:

(a) - 7.1 cm
(b) 0 cm/sec