Math Problem Statement

A 0.51-kg mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface. Determine the following: (a) amplitude A of the motion, (b) angular frequency ω, (c) spring constant k, (d) speed of the object at t = 2.0 s, (e) magnitude of the object's acceleration at t = 2.0 s.

Solution

Let's work through the problem step by step. From the image and question, we are solving for parts (d) and (e):


Part (d): Speed at t=2.0st = 2.0 \, \mathrm{s}

The speed of the object can be found using the derivative of the position function with respect to time, which gives the velocity. The magnitude of velocity at a specific point is the speed:

  1. The position function for simple harmonic motion (SHM) is: x(t)=Acos(ωt)x(t) = A \cos(\omega t) where:

    • A=0.10mA = 0.10 \, \mathrm{m} (amplitude),
    • ω=0.79rad/s\omega = 0.79 \, \mathrm{rad/s} (angular frequency).
  2. The velocity function v(t)v(t) is obtained by differentiating x(t)x(t): v(t)=Aωsin(ωt)v(t) = -A \omega \sin(\omega t)

  3. Substituting A=0.10mA = 0.10 \, \mathrm{m} and ω=0.79rad/s\omega = 0.79 \, \mathrm{rad/s}: v(t)=0.100.79sin(0.79t)v(t) = -0.10 \cdot 0.79 \sin(0.79 \cdot t)

  4. At t=2.0st = 2.0 \, \mathrm{s}: v(2.0)=0.100.79sin(0.792.0)v(2.0) = -0.10 \cdot 0.79 \sin(0.79 \cdot 2.0) Compute the sine term and calculate v(t)v(t).


Part (e): Acceleration at t=2.0st = 2.0 \, \mathrm{s}

The acceleration in SHM is related to the second derivative of the position function or directly to the position function by: a(t)=ω2x(t)a(t) = -\omega^2 x(t)

  1. Substituting ω=0.79rad/s\omega = 0.79 \, \mathrm{rad/s}: a(t)=(0.79)2x(t)a(t) = -(0.79)^2 x(t)

  2. At t=2.0st = 2.0 \, \mathrm{s}, find x(2.0)x(2.0) from: x(t)=Acos(ωt)x(t) = A \cos(\omega t) x(2.0)=0.10cos(0.792.0)x(2.0) = 0.10 \cos(0.79 \cdot 2.0)

  3. Substitute x(2.0)x(2.0) into the acceleration equation to compute a(2.0)a(2.0).


Let me calculate these values for you. math

Given values

A = 0.10 # amplitude in meters omega = 0.79 # angular frequency in rad/s t = 2.0 # time in seconds

Velocity at t = 2.0 s

v_t = -A * omega * math.sin(omega * t)

Position at t = 2.0 s

x_t = A * math.cos(omega * t)

Acceleration at t = 2.0 s

a_t = -(omega**2) * x_t

v_t, a_t### Results:

  1. Speed of the object at t=2.0st = 2.0 \, \mathrm{s}: v(2.0)=0.079m/s|v(2.0)| = 0.079 \, \mathrm{m/s}

  2. Acceleration of the object at t=2.0st = 2.0 \, \mathrm{s}: a(2.0)=0.00057m/s2a(2.0) = 0.00057 \, \mathrm{m/s^2}


Would you like a detailed explanation of these results or assistance with another part of the problem?


Follow-up Questions:

  1. How do sinusoidal functions relate to velocity and acceleration in simple harmonic motion?
  2. What is the physical interpretation of negative velocity and acceleration values in this context?
  3. Can we determine the object's total energy using the given data?
  4. How does the spring constant influence the angular frequency in SHM?
  5. How can we verify the correctness of the computed speed using graphical data?

Tip: The maximum speed in SHM is achieved when the object passes through the equilibrium position, and it's given by vmax=Aωv_\text{max} = A\omega.

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Math Problem Analysis

Mathematical Concepts

Simple Harmonic Motion
Trigonometric Derivatives
Velocity and Acceleration in SHM

Formulas

x(t) = A cos(ωt)
v(t) = -Aω sin(ωt)
a(t) = -ω^2 x(t)
ω = sqrt(k / m)

Theorems

Hooke's Law for Springs
Simple Harmonic Motion equations

Suitable Grade Level

Grades 11-12 / College Physics