Undergraduate · Harmonic oscillators · 368 of 650
Harmonic oscillators · Angular frequency ω=2; Initial displacement b=-1.5
Evaluate at time t=6π/5, retaining π during calculation. Find displacement, velocity, and conserved energy (y′²+ω²y²)/2. Givens: Angular frequency ω=2; Initial displacement b=-1.5.
Try the question, use a hint when you need one, and compare your reasoning with the worked solution. The animation starts with this chapter’s values; changing its controls explores a new case.
01 · Make a prediction
Your practice question
An undamped oscillator satisfies y″+(2)²y=0 with y(0)=-1.5 and y′(0)=1. Mass is normalized to one. Evaluate at time t=6π/5, retaining π during calculation. Find displacement, velocity, and conserved energy (y′²+ω²y²)/2.
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A starting point
The solution is b cos(ωt)+sin(ωt)/ω. Differentiate it before evaluating the velocity.
Work through the reasoning
Step 1
Identify the model and target
An undamped oscillator satisfies y″+(2)²y=0 with y(0)=-1.5 and y′(0)=1. Mass is normalized to one. Evaluate at time t=6π/5, retaining π during calculation. The governing relation is y″+ω²y=0; y=b cos(ωt)+sin(ωt)/ω. The solution is b cos(ωt)+sin(ωt)/ω. Differentiate it before evaluating the velocity.
Step 2
Substitute and calculate
At t=3.769911, y=-1.5cos(2t)+sin(2t)/2=0.012003. Differentiate to obtain y′=3.162187; energy is [(3.162187)²+(2)²(0.012003)²]/2=5.
Step 3
Check the mathematical meaning
The initial energy is [1+(2)²(-1.5)²]/2=5. It must match the energy at the requested time even though position and velocity change. Time: 3.77; Displacement: 0.012; Velocity: 3.162; Conserved energy: 5. Decimal values are rounded, so use unrounded intermediate values.
The answer
At t=3.769911, y=-1.5cos(2t)+sin(2t)/2=0.012003. Differentiate to obtain y′=3.162187; energy is [(3.162187)²+(2)²(0.012003)²]/2=5. The initial energy is [1+(2)²(-1.5)²]/2=5. It must match the energy at the requested time even though position and velocity change. Animation check: Time: 3.77; Displacement: 0.012; Velocity: 3.162; Conserved energy: 5. Decimal displays are rounded; retain the original parameters and exact π until the final step.
Compare the method as well as the result. A different valid method may reach the same answer. Keep exact values until the last step when the question asks for rounding.
02 · Explore the model
See the mathematical relationship move.
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Watch the relationship
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The mathematical idea
A second-order equation needs both initial position and initial velocity. Here the velocity starts at one, so changing frequency also changes the sine coefficient needed to preserve that initial condition. The sum of kinetic and potential quadratic energies stays constant throughout the exact motion. An undamped oscillator satisfies y″+(2)²y=0 with y(0)=-1.5 and y′(0)=1. Mass is normalized to one. Evaluate at time t=6π/5, retaining π during calculation. The requested state occurs at 60% playback; use the exact target stated in the question for your calculation.
y″+ω²y=0; y=b cos(ωt)+sin(ωt)/ω
03 · Reflect and transfer
Explain what changes and why.
If frequency changes while initial velocity stays one, why must the sine coefficient change too?
This is a distinct guided scenario using a reusable mathematical model. Similar-looking diagrams can represent different given values and conclusions; they are not different mathematical theories.