Question

1. An ideal (frictionless) simple harmonic oscillator is set into motion by releasing it from rest at X +0.750 m. The oscillator is set into motion once again from x=+0.750 m, except the oscillator now experiences a retarding force that is linear with respect to velocity. As a result, the oscillator does not return to its original starting position, but instead reaches = +0.700 m after one period. a. During the first full oscillation of motion, determine the fraction of the oscillator's total energy that was dissipated due to the retarding force. Explain. b. Predict the maximum displacement of the oscillator upon completing its fourth full oscillation Explain your reasoning c. Determine the quality factor of the oscillator. Show all work 2. Consider two identical simple harmonic oscillators (1 and 2) to which are applied different damping forces. The two oscillators become underdamped, with the frequency of the first oscillator now greater than that of the second (ie, 0 6 .2). a. Which oscillator has the larger damping constant: oscillator 1, oscillator 2. neither, or is there insufficient information to tell? Explain your reasoning. b. Which oscillator retains the larger fraction of its amplitude with each oscillation: oscillator / oscillator 2, neither, or is there insufficient information to tell? Explain your reasoning, 3. In section Il of the tutorial we found the following expression for the quality factor: 0 - 1-es In this problem we consider the case in which the quantity YT is much smaller than one ( 1) a. Show that such an oscillator is not only underdamped but weakly damped, ie, that the damping constant is much smaller than the natural frequency of the oscillator: y el b. Extend your results by showing that the quality factor of a weakly damped oscillator can be approximated as w2y where ay is the angular frequency of the oscillator Hint: Use the power series expansion of the exponential function: '-1 + x + + +... 4. Consider a weakly damped oscillator that is released from rest at 1 = 0 and whose amplitude is smaller by a factor of e after undergoing N full oscillations a. Write an expression for the period of oscillation Ta in terms of N and the damping constant Explain your reasoning. (Hint: Recall your interpretation of the quantity e ) b. The quality factor of this oscillator might be proportional to N. inversely proportional to Nor independent of N. Which of these options makes the most sense? Explain qualitatively, without performing any calculations or referring to any formulas. c. Verify your answer in part babove finding an expression for in terms of N. Show all work.

Answer 1

(a) initial energy stored in oscillator = (1/2)k x_o² = (1/2)k(0.75)²

where k is spring constant, x_o is initial amplitude

energy of oscillator after one oscillation (1/2)k(0.7)²

energy dissipated in one oscillation = (1/2)k [ ( 0.75 )² - (0.70)² ]

fraction of energy dissipated in one oscillation = [ (0.75)² - (0.7)²] / [ (0.75)² ] = 0.129

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amplitude of damped oscillator is given by, A = A_o e^{- ($\gamma$t ) / 2}

If initial amplitude A_o = 0.75 m and amplitude A after one oscillation is 0.7 m, then we have

0.7 = 0.75 e^{- ( $\gamma$ T ) / 2}   or e^{- ( $\gamma$ T ) / 2} = ( 0.70/0.75 ) = 0.933 ...............(1)

after 4 oscillation, amplitudes ratio e^{- ( $\gamma$ 4T ) / 2} = ( e^{- ( $\gamma$ T ) / 2} )⁴ = ( 0.933 )⁴ = 0.759

hence amplitude after 4th oscillation = 0.75 $\times$ 0.759 $\approx$ 0.57 m

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Quality factor Q = 2$\pi$ ( E /$\Delta$E )

where E is initial energy and $\Delta$ E is energy loss in one oscillation

we have the fraction ($\Delta$E / E ) from part-1 as, ($\Delta$E / E ) = 0.129

Hence quality factor Q = 2$\pi$ / 0.129 = 48.71