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Derive the frequency of oscillation of a torsional oscillator, Equation 11.20. Hint: Consider how the restoring force depends on the twist angle and use Newton's second law for rotational motion (7 = Ia).

Derive the frequency of oscillation of a torsional oscillator, Equation 11.20. Hint: Consider how the restoring force depends on the twist angle and use Newton's second law for rotational motion (7 = Ia).

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Question 14. Must use formula from formula sheet. Not sure which one
11. Make a sketch of how the position and kinetic energy of a harmonic oscillator vary with time. The period of the oscillations of KE can be determined from the separation in time of adjacent maxima in your KE sketch. Show that the frequency of the KE (2) oscillations is equal to twice the frequency of the displacement oscillations (Eq. 11.14). Explain why. 12. In Section 11.3, we discussed the total mechanical energy of a mass-on-a-spring oscillator. The result in Equation 11.21 shows that the total energy is proportional to the square of the ampli- E en tude. Derive the corresponding results for the total mechani- cal energy of a simple pendulum and for a torsional oscillator 32. and show that they are also proportional to the square of the ri amplitude. शीत श 13. SSM Use energy considerations to derive the oscillation fre- quency for a mass-on-a-spring oscillator. Hint: The maximum potential energy stored in the spring must be equal to the maxi- mum kinetic energy of the mass. 14. Derive the frequency of oscillation of a torsional oscillator, Equation 11.20. Hint: Consider how the restoring force depends on the twist angle and use Newton's second law for rotational 33. motion (7 = la). 15. Pendulum clocks use a pendulum to keep time (Fig. Q11.15). This type of clock can be adjusted by varying the length of the pendulum. How should the length be adjusted 34. in the following cases? (a) The clock is running slow. (b) The clock is running fast. (c) The clock is running fine at sea level, but needs to
Transcribed Image Text:11. Make a sketch of how the position and kinetic energy of a harmonic oscillator vary with time. The period of the oscillations of KE can be determined from the separation in time of adjacent maxima in your KE sketch. Show that the frequency of the KE (2) oscillations is equal to twice the frequency of the displacement oscillations (Eq. 11.14). Explain why. 12. In Section 11.3, we discussed the total mechanical energy of a mass-on-a-spring oscillator. The result in Equation 11.21 shows that the total energy is proportional to the square of the ampli- E en tude. Derive the corresponding results for the total mechani- cal energy of a simple pendulum and for a torsional oscillator 32. and show that they are also proportional to the square of the ri amplitude. शीत श 13. SSM Use energy considerations to derive the oscillation fre- quency for a mass-on-a-spring oscillator. Hint: The maximum potential energy stored in the spring must be equal to the maxi- mum kinetic energy of the mass. 14. Derive the frequency of oscillation of a torsional oscillator, Equation 11.20. Hint: Consider how the restoring force depends on the twist angle and use Newton's second law for rotational 33. motion (7 = la). 15. Pendulum clocks use a pendulum to keep time (Fig. Q11.15). This type of clock can be adjusted by varying the length of the pendulum. How should the length be adjusted 34. in the following cases? (a) The clock is running slow. (b) The clock is running fast. (c) The clock is running fine at sea level, but needs to
Physics 104 Equation Sheet k = 9.0 x 10° Nm²/C² Ho = 4n x 107 N/A? C = 3.00 x 108 m/s e = 1.6 x 1019 C E0 = 8.85 x 1012 C²/Nm2 h = 6.63 x 10 ³34 Js = 4,14 x 10 15 eVs lo = 1012 W/m me = 9.11 x 10 31 kg m, = 1.67 x 1027 kg nair = 1.00 Nwater = 1.33 nsoap = 1.50 Vsound = 340 m/s Wsiltver = 4.74 eV 1 eV = 1.6 x 10-19 J F = kqıq2/r? E = kQ/r? F = qE C = q/AV AV = IR R = pL/A F = ILB sino PE = kqıq2/r V = kQ/r AV = -EAx APEE = -WE = -qEd APE = qAV C = E0A/d PE = ½ QAV Reg = E; R, 1/Req Σ 1/R I= Aq/At F = qvB sino EMF = NA D8/At P = IAV B = Hol/2nr B = Ho nl OB = BA cosO IV1 = 12 V2 V1/V2 = N1/N2 EMF = NBAW sin(wt) v = fA n = c/v f = 1/T W = 2nf = F = -kAx PE = ½ kAx? An = Aair/n V = fs fo = ni sino1 = n2 sin©2 B = 10 log10 (1/lo) 1/o + 1/i = 1/f m = -i/o f = R/2 P = 1/f fo = fs(1 ± Po/p) d sino = (m + ½ )A d sino = mA E = hf Am T = 2.90 x 103 mK KEmax = hf – W En = -13.6eV/n² Rn = (0.0529 nm) n? A = h/mv = h/p A-X = (h/mc)(1– cos©) L= nh/2n AxAp, 2 h/4n ΔΕΔΕ h/4π
Transcribed Image Text:Physics 104 Equation Sheet k = 9.0 x 10° Nm²/C² Ho = 4n x 107 N/A? C = 3.00 x 108 m/s e = 1.6 x 1019 C E0 = 8.85 x 1012 C²/Nm2 h = 6.63 x 10 ³34 Js = 4,14 x 10 15 eVs lo = 1012 W/m me = 9.11 x 10 31 kg m, = 1.67 x 1027 kg nair = 1.00 Nwater = 1.33 nsoap = 1.50 Vsound = 340 m/s Wsiltver = 4.74 eV 1 eV = 1.6 x 10-19 J F = kqıq2/r? E = kQ/r? F = qE C = q/AV AV = IR R = pL/A F = ILB sino PE = kqıq2/r V = kQ/r AV = -EAx APEE = -WE = -qEd APE = qAV C = E0A/d PE = ½ QAV Reg = E; R, 1/Req Σ 1/R I= Aq/At F = qvB sino EMF = NA D8/At P = IAV B = Hol/2nr B = Ho nl OB = BA cosO IV1 = 12 V2 V1/V2 = N1/N2 EMF = NBAW sin(wt) v = fA n = c/v f = 1/T W = 2nf = F = -kAx PE = ½ kAx? An = Aair/n V = fs fo = ni sino1 = n2 sin©2 B = 10 log10 (1/lo) 1/o + 1/i = 1/f m = -i/o f = R/2 P = 1/f fo = fs(1 ± Po/p) d sino = (m + ½ )A d sino = mA E = hf Am T = 2.90 x 103 mK KEmax = hf – W En = -13.6eV/n² Rn = (0.0529 nm) n? A = h/mv = h/p A-X = (h/mc)(1– cos©) L= nh/2n AxAp, 2 h/4n ΔΕΔΕ h/4π
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