If the amplitude of a damped oscillator decreases to half its value in 5 seconds, what is the damping ratio?

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Q: If the amplitude of a damped oscillator decreases to half its value in 5 seconds, what is the damping ratio?

Solution: Using the formula A(t) = A_0 e^(-ζω_nt), we find ζ = 0.2.

Steps: 9

Step 1: Understand the formula A(t) = A_0 e^(-ζω_nt), where A(t) is the amplitude at time t, A_0 is the initial amplitude, ζ is the damping ratio, and ω_n is the natural frequency.
Step 2: Recognize that the problem states the amplitude decreases to half its value in 5 seconds. This means A(5) = 0.5 * A_0.
Step 3: Substitute A(5) into the formula: 0.5 * A_0 = A_0 e^(-ζω_n * 5).
Step 4: Divide both sides by A_0 (assuming A_0 is not zero): 0.5 = e^(-ζω_n * 5).
Step 5: Take the natural logarithm of both sides: ln(0.5) = -ζω_n * 5.
Step 6: Solve for ζ: ζ = -ln(0.5) / (5 * ω_n).
Step 7: Since we need the damping ratio, we need to know ω_n. However, if we assume ω_n = 1 for simplicity, we can calculate ζ = -ln(0.5) / 5.
Step 8: Calculate ln(0.5) which is approximately -0.693. Then, ζ = 0.693 / 5 = 0.1386.
Step 9: If we assume ω_n is 1, we find ζ ≈ 0.1386. If ω_n is different, adjust accordingly.