a weight is attached to a spring and reaches its equilibrium position (x = 0). it is then set in motion…

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)\n-7.1 cm\n(type an integer or decimal rounded to one decimal place as needed.)\n(b) what is the springs velocity when t = 0?\n0 cm/sec\n(type an integer or decimal rounded to one decimal place as needed.)\nwhat is the springs velocity when t = π/3?\n cm/sec\n(type an integer or decimal rounded to one decimal place as needed.)

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)\n-7.1 cm\n(type an integer or decimal rounded to one decimal place as needed.)\n(b) what is the springs velocity when t = 0?\n0 cm/sec\n(type an integer or decimal rounded to one decimal place as needed.)\nwhat is the springs velocity when t = π/3?\n cm/sec\n(type an integer or decimal rounded to one decimal place as needed.)

Answer

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 $v=\frac{dx}{dt}$.

Step2: Differentiate the displacement function

Using the derivative formula $\frac{d}{dt}(\cos t)=-\sin t$, we have $v = - 10\sin t$.

Step3: Calculate velocity at $t = \frac{\pi}{3}$

Substitute $t=\frac{\pi}{3}$ into the velocity - function $v=-10\sin t$. Since $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$, then $v=-10\times\frac{\sqrt{3}}{2}=- 5\sqrt{3}\approx - 8.7$.

Answer:

$-8.7$