Questions: 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=6 cos t, where x is measured in centimeters and t is measured in seconds. Answer parts (a) and (b). (b) What is the spring's velocity when t=0 ? What is the spring's velocity when t=pi/3 ? What is the spring's velocity when t=3 pi/4 ?

Solution Steps

• The displacement of the spring is given by \( x(t) = 6 \cos t \).
• The velocity \( v(t) \) is the derivative of the displacement with respect to time \( t \).
• Differentiate \( x(t) \) to find \( v(t) \): \[ v(t) = \frac{d}{dt}[6 \cos t] = -6 \sin t \]

• Substitute \( t = 0 \) into the velocity function: \[ v(0) = -6 \sin(0) = 0 \]

• Substitute \( t = \frac{\pi}{3} \) into the velocity function: \[ v\left(\frac{\pi}{3}\right) = -6 \sin\left(\frac{\pi}{3}\right) = -6 \times \frac{\sqrt{3}}{2} = -3\sqrt{3} \]
• Approximate the value: \[ -3\sqrt{3} \approx -5.2 \]

• Substitute \( t = \frac{3\pi}{4} \) into the velocity function: \[ v\left(\frac{3\pi}{4}\right) = -6 \sin\left(\frac{3\pi}{4}\right) = -6 \times \frac{\sqrt{2}}{2} = -3\sqrt{2} \]
• Approximate the value: \[ -3\sqrt{2} \approx -4.2 \]

Final Answer