Math Problem Statement

Determine expressions for maximum position, velocity, and acceleration, and where in the cycle these values occur. Use Newton's Second Law to find maximum acceleration and angular frequency, and calculate the period of the mass.

Solution

I see the worksheet you've provided involves problems requiring mathematical modeling and analysis based on physics equations, particularly related to motion and forces. Here's an explanation and solution approach for each question visible:


Question 5:

Task: Determine expressions for ymax,vmax,y_{\text{max}}, v_{\text{max}}, and amaxa_{\text{max}}. Analyze where these values occur in the cycle.

  • Given: Equation y=0.003479cos(8.143t+5.667)+0.4594y = 0.003479 \cos(8.143t + 5.667) + 0.4594.

Steps:

  1. ymaxy_{\text{max}} occurs when cos(8.143t+5.667)=1\cos(8.143t + 5.667) = 1.

    • Maximum value: ymax=0.0034791+0.4594=0.462879y_{\text{max}} = 0.003479 \cdot 1 + 0.4594 = 0.462879.
  2. vmaxv_{\text{max}}:

    • Velocity v(t)=dydtv(t) = \frac{dy}{dt}.
    • From the equation v(t)=0.028329sin(8.143t+5.667)v(t) = -0.028329 \sin(8.143t + 5.667), the maximum value occurs when sin(8.143t+5.667)=±1\sin(8.143t + 5.667) = \pm 1:
    • vmax=0.028329v_{\text{max}} = 0.028329.
  3. amaxa_{\text{max}}:

    • Acceleration a(t)=d2ydt2a(t) = \frac{d^2y}{dt^2}.
    • From the equation a(t)=0.230697cos(8.143t+5.667)a(t) = -0.230697 \cos(8.143t + 5.667), the maximum value occurs when cos(8.143t+5.667)=±1\cos(8.143t + 5.667) = \pm 1:
    • amax=0.230697a_{\text{max}} = 0.230697.

Question 6:

Task: Write an equation for maximum acceleration and calculate angular frequency ω\omega using Newton's Second Law.

  • Newton's Second Law: F=maF = ma, where a=ω2Aa = \omega^2 A.
  • Maximum acceleration: amax=ω2Aa_{\text{max}} = \omega^2 A.
    • Using A=0.003479A = 0.003479 and amax=0.230697a_{\text{max}} = 0.230697:
      • ω2=amaxA=0.2306970.00347966.3\omega^2 = \frac{a_{\text{max}}}{A} = \frac{0.230697}{0.003479} \approx 66.3.
      • ω=66.38.143 rad/s\omega = \sqrt{66.3} \approx 8.143 \ \text{rad/s} (consistent with given data).

Question 7:

Task: Calculate the period TT using T=2πωT = \frac{2\pi}{\omega}.

  • Given: ω=8.143 rad/s\omega = 8.143 \ \text{rad/s}.
  • Period: T=2π8.1430.771 sT = \frac{2\pi}{8.143} \approx 0.771 \ \text{s}.

Comparison with Graphs:

  • Check if the period matches the graph's periodicity. If consistent, this validates the model.

Would you like detailed derivations, or do you have further questions? Here are some relative queries:

  1. How is the period TT connected to angular frequency ω\omega?
  2. What is the significance of finding ymax,vmax,amaxy_{\text{max}}, v_{\text{max}}, a_{\text{max}}?
  3. How does Newton's Second Law apply to harmonic motion?
  4. What would happen to the period if ω\omega were doubled?
  5. How do phase shifts (e.g., +5.667+5.667) affect the motion's maximum values?

Tip: Always relate angular frequency ω\omega to measurable parameters like period TT and maximum velocity/acceleration for deeper insights.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometry
Calculus
Physics

Formulas

y = Acos(ωt + φ)
v(t) = -Aωsin(ωt + φ)
a(t) = -Aω^2cos(ωt + φ)
T = 2π/ω
F = ma

Theorems

Newton's Second Law
Simple Harmonic Motion

Suitable Grade Level

Grades 11-12, College Physics