Lightly damped oscillator: log decrement, Q, and two notebook claims
A student in a teaching laboratory suspends a 200 g mass from a vertical spring of stiffness 20 N/m and sets it oscillating vertically. Using a motion sensor she records the positions of successive maxima on the same side of the equilibrium point: 8.0 cm, 5.9 cm and 4.3 cm, with a time gap of 0.63 s between neighbouring maxima. Readings are quoted to two significant figures. The spring-mass system is expected to be only lightly damped. In her notebook she writes two statements: (i) "the amplitude falls to half in three oscillations"; (ii) "since Q is about 10, the energy stored in the oscillator drops to 1/e of its initial value after about ten oscillations". The demonstrator asks her to check both statements quantitatively before submitting the report, and to state clearly which approximations she is implicitly using when she treats the motion as simple harmonic motion.
1. From the amplitude data and the period, obtain the logarithmic decrement, the damping constant gamma and the quality factor Q. Justify why the observed period may be identified with the undamped period.
Read the worked answer
Successive maxima on the same side are one period apart, so T = 0.63 s. The logarithmic decrement is delta = ln(8.0/5.9) = 0.305, and the second ratio gives ln(5.9/4.3) = 0.316; average delta = 0.31.
Damping constant: gamma = delta/T = 0.31/0.63 = 0.49 s^-1.
Check the undamped frequency: omega_0 = sqrt(k/m) = sqrt(20/0.200) = 10 rad/s, so T_0 = 2*pi/10 = 0.628 s. The measured T = 0.63 s agrees with T_0 to the quoted precision. Since the damped period is T = T_0/sqrt(1 - (gamma/omega_0)^2) and gamma/omega_0 = 0.049, the correction is about 0.1 per cent - negligible. That is exactly the justification for identifying T with T_0.
Quality factor: Q = omega_0/(2*gamma) = 10/(0.98) = 10.2, consistent with Q = pi/delta = 10.1.
2. Test statement (i): does the amplitude fall to half in three oscillations? Support your answer with numbers.
Read the worked answer
The envelope decays as A_n = A_0 exp(-n*delta). After three periods, A_3 = 8.0 * exp(-3 x 0.31) = 8.0 x 0.395 = 3.16 cm. Halving 8.0 cm would give 4.0 cm, which is not observed; the measured third maximum is 4.3 cm, and the exponential prediction 3.16 cm differs from that by more than the reading scatter, so the exponential model is itself only approximate here.
The correct statement is obtained by solving 0.5 = exp(-n*delta): n = ln2/delta = 0.693/0.31 = 2.2 periods. So the amplitude halves in about two oscillations, not three.
The judgement call is that a factor-of-two statement must be checked against the decay law rather than guessed from the visual size of the decrement; three periods correspond to a factor exp(-0.93) = 0.39, not 0.5.
3. Test statement (ii) about the energy decaying to 1/e in Q oscillations. What is the correct interpretation of Q = 10 here?
Read the worked answer
Energy is proportional to amplitude squared, so E(t) = E_0 exp(-2gammat). Setting 2gammat = 1 gives t = 1/(2 x 0.49) = 1.0 s, i.e. about 1.6 periods - nowhere near ten oscillations.
The meaning of Q is the fractional energy loss per radian: Q = 2pi(energy stored)/(energy lost per cycle). With Q = 10 the fractional energy lost per cycle is 2pi/Q = 0.63, that is about 63 per cent per oscillation. Over Q oscillations the surviving fraction is exp(-2pi) = 1.9 x 10^-3, not 1/e.
So statement (ii) is a units/interpretation error: Q is not the number of oscillations for 1/e energy decay. The candidate should recognise that a Q of only 10 is quite strongly damped for this purpose, even though the amplitude decay per period looks small.