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A 0.830-kg block oscillates on the end of a spring whose spring constant is k = 42.0 N/m . The Part B mass moves in a fluid which offers a resistive force F = -bv, where b = 0.630 N · s/m. What is the fractional decrease in amplitude per cycle? ΑΣφ ? ΔΑ %3D A Submit Request Answer

A 0.830-kg block oscillates on the end of a spring whose spring constant is k = 42.0 N/m . The Part B mass moves in a fluid which offers a resistive force F = -bv, where b = 0.630 N · s/m. What is the fractional decrease in amplitude per cycle? ΑΣφ ? ΔΑ %3D A Submit Request Answer

Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter12: Oscillatory Motion
Section: Chapter Questions
Problem 66P: Consider the damped oscillator illustrated in Figure 12.16a. The mass of the object is 375 g, the...
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Need help with part Band C pls. For part A, i found that the period is T= 0.88327 s

Constants I Perioc A 0.830-kg block oscillates on the end of a spring whose spring constant is k = 42.0 N/m . The Part B mass moves in a fluid which offers a resistive force F = -bv, where b = 0.630 N · s/m. What is the fractional decrease in amplitude per cycle? ν ΑΣφ ? ΔΑ A Submit Request Answer Part C Write the displacement x as a function of time t in seconds if at t = 0, x = 0, and at t = 1.00s, x = 0.120m. x = (0.240m) sin[(7.10rad/s)t] x = (0.240m)e-(0.380s")t sin[(7.10rad/s)t] x = (0.240m)e(0.3805-)t cos[(7.10rad/s)t] O x = (0.240m) cos[(7.10rad/s)t] Submit Request Answer
Transcribed Image Text:Constants I Perioc A 0.830-kg block oscillates on the end of a spring whose spring constant is k = 42.0 N/m . The Part B mass moves in a fluid which offers a resistive force F = -bv, where b = 0.630 N · s/m. What is the fractional decrease in amplitude per cycle? ν ΑΣφ ? ΔΑ A Submit Request Answer Part C Write the displacement x as a function of time t in seconds if at t = 0, x = 0, and at t = 1.00s, x = 0.120m. x = (0.240m) sin[(7.10rad/s)t] x = (0.240m)e-(0.380s")t sin[(7.10rad/s)t] x = (0.240m)e(0.3805-)t cos[(7.10rad/s)t] O x = (0.240m) cos[(7.10rad/s)t] Submit Request Answer
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