Math Problem Statement

A mass attached to a spring executes simple harmonic motion in a horizontal plane with an amplitude of 0.382 m. At a point 0.14134 m away from the equilibrium, the mass has speed 0.487 m/s. What is the period of oscillation of the mass? Consider equations for xt) and vt) and use sin? + cos? = 1 to calculate w. Answer in units of s.

Solution

To solve for the period of oscillation TT of a mass undergoing simple harmonic motion, we’ll start by applying the equations for displacement x(t)x(t) and velocity v(t)v(t) of a simple harmonic oscillator.

Given Data:

  • Amplitude, A=0.382mA = 0.382 \, \text{m}
  • Displacement from equilibrium, x=0.14134mx = 0.14134 \, \text{m}
  • Speed at this displacement, v=0.487m/sv = 0.487 \, \text{m/s}

Step 1: Set up the equations

For simple harmonic motion:

  1. The displacement x(t)x(t) is given by: x(t)=Acos(ωt)x(t) = A \cos(\omega t)
  2. The velocity v(t)v(t) is given by: v(t)=Aωsin(ωt)v(t) = -A \omega \sin(\omega t) where ω\omega is the angular frequency.

Step 2: Use the identity sin2(ωt)+cos2(ωt)=1\sin^2(\omega t) + \cos^2(\omega t) = 1

From x(t)=Acos(ωt)x(t) = A \cos(\omega t), we have: cos(ωt)=xA\cos(\omega t) = \frac{x}{A} Substituting the values: cos(ωt)=0.141340.382\cos(\omega t) = \frac{0.14134}{0.382} Calculating this gives: cos(ωt)0.37\cos(\omega t) \approx 0.37

Using sin2(ωt)+cos2(ωt)=1\sin^2(\omega t) + \cos^2(\omega t) = 1, we find sin(ωt)\sin(\omega t): sin(ωt)=1cos2(ωt)=1(0.37)2\sin(\omega t) = \sqrt{1 - \cos^2(\omega t)} = \sqrt{1 - (0.37)^2} sin(ωt)0.9295\sin(\omega t) \approx 0.9295

Step 3: Solve for ω\omega using the velocity equation

We have v(t)=Aωsin(ωt)v(t) = -A \omega \sin(\omega t), so: ω=vAsin(ωt)\omega = \frac{v}{A \sin(\omega t)} Substitute the known values: ω=0.4870.382×0.9295\omega = \frac{0.487}{0.382 \times 0.9295} Calculating this gives: ω1.37rad/s\omega \approx 1.37 \, \text{rad/s}

Step 4: Find the period TT

The period TT is given by: T=2πωT = \frac{2\pi}{\omega} Substituting ω1.37rad/s\omega \approx 1.37 \, \text{rad/s}: T2π1.374.59sT \approx \frac{2 \pi}{1.37} \approx 4.59 \, \text{s}

Final Answer:

The period of oscillation TT is approximately 4.59s4.59 \, \text{s}.

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Related Questions:

  1. What is the maximum speed of the mass during its motion?
  2. How does the amplitude of oscillation affect the period?
  3. What is the acceleration of the mass at a displacement of 0.14134 m?
  4. If the amplitude were doubled, how would the period change?
  5. How is the energy distributed between kinetic and potential energy at a displacement of 0.14134 m?

Tip: In simple harmonic motion, the period depends only on mass and spring constants (for spring systems) or length and gravity (for pendulums), not on amplitude.

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Math Problem Analysis

Mathematical Concepts

Simple Harmonic Motion
Oscillations
Angular Frequency
Trigonometric Identities

Formulas

x(t) = A cos(ωt)
v(t) = -Aω sin(ωt)
ω = v / (A sin(ωt))
T = 2π / ω
sin^2(ωt) + cos^2(ωt) = 1

Theorems

Trigonometric Identity

Suitable Grade Level

Grades 11-12