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 of a mass undergoing simple harmonic motion, we’ll start by applying the equations for displacement and velocity of a simple harmonic oscillator.
Given Data:
- Amplitude,
- Displacement from equilibrium,
- Speed at this displacement,
Step 1: Set up the equations
For simple harmonic motion:
- The displacement is given by:
- The velocity is given by: where is the angular frequency.
Step 2: Use the identity
From , we have: Substituting the values: Calculating this gives:
Using , we find :
Step 3: Solve for using the velocity equation
We have , so: Substitute the known values: Calculating this gives:
Step 4: Find the period
The period is given by: Substituting :
Final Answer:
The period of oscillation is approximately .
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Related Questions:
- What is the maximum speed of the mass during its motion?
- How does the amplitude of oscillation affect the period?
- What is the acceleration of the mass at a displacement of 0.14134 m?
- If the amplitude were doubled, how would the period change?
- 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
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