Question

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--.

Explanation:

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 \(a(t)=\frac{dv}{dt}=-A\sin(t)\).

1. (a) Expressions for \(v(t)\) and \(a(t)\)

• \(v(t)=A\cos(t)\)
• \(a(t)= - A\sin(t)\)

1. (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 \(a(\frac{2\pi}{3})=-A\sin(\frac{2\pi}{3})=-\frac{\sqrt{3}}{2}A\).

1. 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.