a mass on a spring vibrates horizontally on a smooth, level surface as shown in the figure. its equation of…

a mass on a spring vibrates horizontally on a smooth, level surface as shown in the figure. its equation of motion is x(t)=8 sin(t), where t is measured in seconds and x in centimeters. (a) find the velocity and acceleration at time t. v(t)= a(t)= (b) find the position, velocity, and acceleration of the mass at time t = 2π/3. x(2π/3)= v(2π/3)= a(2π/3)= in what direction is the mass moving at that time? since v(2π/3)? 0, the mass is moving to the --select--.

a mass on a spring vibrates horizontally on a smooth, level surface as shown in the figure. its equation of motion is x(t)=8 sin(t), where t is measured in seconds and x in centimeters. (a) find the velocity and acceleration at time t. v(t)= a(t)= (b) find the position, velocity, and acceleration of the mass at time t = 2π/3. x(2π/3)= v(2π/3)= a(2π/3)= in what direction is the mass moving at that time? since v(2π/3)? 0, the mass is moving to the --select--.

Answer

  1. First, recall the relationships for a mass - spring system with the position function (x(t)=A\sin(t)):
    • The velocity function (v(t)) is the derivative of the position function, and the acceleration function (a(t)) is the derivative of the velocity function.
    • Using the derivative rules, if (x(t) = A\sin(t)), then:
      • Step 1: Find the velocity function
        • The derivative of (y = \sin(t)) with respect to (t) is (y^\prime=\cos(t)). So, if (x(t)=A\sin(t)), then (v(t)=\frac{dx}{dt}=A\cos(t)).
      • Step 2: Find the acceleration function
        • The derivative of (y = \cos(t)) with respect to (t) is (y^\prime=-\sin(t)). So, if (v(t)=A\cos(t)), then (a(t)=\frac{dv}{dt}=-A\sin(t)).
  2. (a) Expressions for (v(t)) and (a(t))
    • (v(t)=A\cos(t))
    • (a(t)= - A\sin(t))
  3. (b) Evaluate (x(\frac{2\pi}{3})), (v(\frac{2\pi}{3})), and (a(\frac{2\pi}{3}))
    • Step 1: Evaluate (x(\frac{2\pi}{3}))
      • Substitute (t = \frac{2\pi}{3}) into (x(t)=A\sin(t)). We know that (\sin(\frac{2\pi}{3})=\frac{\sqrt{3}}{2}), so (x(\frac{2\pi}{3})=A\sin(\frac{2\pi}{3})=\frac{\sqrt{3}}{2}A).
    • Step 2: Evaluate (v(\frac{2\pi}{3}))
      • Substitute (t=\frac{2\pi}{3}) into (v(t)=A\cos(t)). We know that (\cos(\frac{2\pi}{3})=-\frac{1}{2}), so (v(\frac{2\pi}{3})=A\cos(\frac{2\pi}{3})=-\frac{1}{2}A).
    • Step 3: Evaluate (a(\frac{2\pi}{3}))
      • Substitute (t = \frac{2\pi}{3}) into (a(t)=-A\sin(t)). Since (\sin(\frac{2\pi}{3})=\frac{\sqrt{3}}{2}), then (a(\frac{2\pi}{3})=-A\sin(\frac{2\pi}{3})=-\frac{\sqrt{3}}{2}A).
  4. Determine the direction of motion when (v(\frac{2\pi}{3}))
    • Since (v(\frac{2\pi}{3})=-\frac{1}{2}A<0) (assuming (A > 0)), the mass is moving to the left.

Answer:

(a) (v(t)=A\cos(t)), (a(t)=-A\sin(t)) (b) (x(\frac{2\pi}{3})=\frac{\sqrt{3}}{2}A), (v(\frac{2\pi}{3})=-\frac{1}{2}A), (a(\frac{2\pi}{3})=-\frac{\sqrt{3}}{2}A), left.