HiggsHunter Posted October 28, 2012 Report Share Posted October 28, 2012 Another question for ya IBS!I thought that a pipe closed at one end only had the first harmonic, third harmonic, fifth harmonic and so on... How would you do this one?I would assume that by "second harmonic frequency" the question is referring to the next higher resonant frequency, which would indeed be three times that of the fundamental. 2 Reply Link to post Share on other sites More sharing options...
eelnedross Posted October 29, 2012 Report Share Posted October 29, 2012 Another question for ya IBS!I thought that a pipe closed at one end only had the first harmonic, third harmonic, fifth harmonic and so on... How would you do this one?I would assume that by "second harmonic frequency" the question is referring to the next higher resonant frequency, which would indeed be three times that of the fundamental.So the third harmonic would have a resonant frequency five times that of the fundamental?And what does it mean exactly with "resonant frequency"? The natural frequency of oscillation which gives the highest amplitude for the oscillation, if i'm not mistaken?Thank you as always Mr. Higgs! Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 29, 2012 Report Share Posted October 29, 2012 Another question for ya IBS!I thought that a pipe closed at one end only had the first harmonic, third harmonic, fifth harmonic and so on... How would you do this one?I would assume that by "second harmonic frequency" the question is referring to the next higher resonant frequency, which would indeed be three times that of the fundamental.So the third harmonic would have a resonant frequency five times that of the fundamental?And what does it mean exactly with "resonant frequency"? The natural frequency of oscillation which gives the highest amplitude for the oscillation, if i'm not mistaken?Thank you as always Mr. Higgs!Yes, it is much more common practice to number the harmonics by the corresponding multiple of the fundamental frequency. But the question appears to adopt the less usual nomenclature of Tsokos (see Q3 on page 254) in which the "second harmonic" means the next after the fundamental. 1 Reply Link to post Share on other sites More sharing options...
eelnedross Posted October 29, 2012 Report Share Posted October 29, 2012 Another question for ya IBS!I thought that a pipe closed at one end only had the first harmonic, third harmonic, fifth harmonic and so on... How would you do this one?I would assume that by "second harmonic frequency" the question is referring to the next higher resonant frequency, which would indeed be three times that of the fundamental.So the third harmonic would have a resonant frequency five times that of the fundamental?And what does it mean exactly with "resonant frequency"? The natural frequency of oscillation which gives the highest amplitude for the oscillation, if i'm not mistaken?Thank you as always Mr. Higgs!Yes, it is much more common practice to number the harmonics by the corresponding multiple of the fundamental frequency. But the question appears to adopt the less usual nomenclature of Tsokos (see Q3 on page 254) in which the "second harmonic" means the next after the fundamental.So for this question we would have to draw 1N - 1A - 1N - 1A in the second diagram?And in a pipe closed at one end, say, if there is a N - A (fundamental frequency), the second harmonic would be the one that has another N - A and so on? In contrast, in a pipe open at both ends, say, if there is a N - N (fundamental frequency), the second harmonic would be the one that has another N - N and so on? I'm not so sure if you understand me. N: NodeA: Antinode Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 29, 2012 Report Share Posted October 29, 2012 Another question for ya IBS!I thought that a pipe closed at one end only had the first harmonic, third harmonic, fifth harmonic and so on... How would you do this one?I would assume that by "second harmonic frequency" the question is referring to the next higher resonant frequency, which would indeed be three times that of the fundamental.So the third harmonic would have a resonant frequency five times that of the fundamental?And what does it mean exactly with "resonant frequency"? The natural frequency of oscillation which gives the highest amplitude for the oscillation, if i'm not mistaken?Thank you as always Mr. Higgs!Yes, it is much more common practice to number the harmonics by the corresponding multiple of the fundamental frequency. But the question appears to adopt the less usual nomenclature of Tsokos (see Q3 on page 254) in which the "second harmonic" means the next after the fundamental.So for this question we would have to draw 1N - 1A - 1N - 1A in the second diagram?And in a pipe closed at one end, say, if there is a N - A (fundamental frequency), the second harmonic would be the one that has another N - A and so on? In contrast, in a pipe open at both ends, say, if there is a N - N (fundamental frequency), the second harmonic would be the one that has another N - N and so on? I'm not so sure if you understand me.N: NodeA: AntinodeYes, except that the fundamental in the pipe open at both ends would be A - N - A (not N - N) and each higher harmonic would have an additional N - A. Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted October 29, 2012 Report Share Posted October 29, 2012 Predicted Question :Why the Transformer frequency must be constant ? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 29, 2012 Report Share Posted October 29, 2012 Predicted Question :Why the Transformer frequency must be constant ?Could you give details of the context of the question? Many transformers (in modulators or amplifiers, for example) are not limited to a single constant frequency. Reply Link to post Share on other sites More sharing options...
eelnedross Posted November 2, 2012 Report Share Posted November 2, 2012 Hey, can someone outline the nature of spectroscopic binary stars and eclipsing binary stars? I can't explain them fully...Also, how can we determine the composition of a star using absorption spectra? I'm lost on that one, and when it includes the Doppler shift! Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted November 2, 2012 Report Share Posted November 2, 2012 Hey, can someone outline the nature of spectroscopic binary stars and eclipsing binary stars? I can't explain them fully...Pairs of binary stars orbit their common centre of mass. The absorption lines in the spectrum of the light received from spectroscopic binaries are red-shifted when a star is moving away from the earth, and blue-shifted when it is moving towards us. The orientation of the orbital plane of eclipsing binaries results in their periodically blocking each other when viewed from earth, modulating their apparent brightness. 1 Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted November 2, 2012 Report Share Posted November 2, 2012 Also, how can we determine the composition of a star using absorption spectra? I'm lost on that one, and when it includes the Doppler shift!The absorption lines in the spectrum of light from a star can indicate the elements present in its atmosphere, providing the temperature of the star is taken into account. The lines are red shifted if the star is receding, and because of its rotation the Doppler shifts can be different for the part moving towards us and the part that is moving away. But since there are many absorption lines, and their patterns are characteristic, these spectral shifts do not normally hinder their identification. 1 Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted November 2, 2012 Report Share Posted November 2, 2012 Question : a) Ionized hydrogen atoms are accelerated from rest in the vacuum between two vertical parallel conducting plates . the potential differnce between the plates is V as a result of accelertation each ion gain energy of 1.9*10^-18 J find the value of V ) b) the plates in (a) are replaced by a cell that has an emf of 12 V and internal resistance 5 ohms . A resistor R is connected in series with the cell . The energy transferred by the cell to an electron as it moves through the resistor is 1.44*10^-18 JShow that the value of R is 15.0 Ohms . I Know how to find a but how can it show the answer to b ?? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted November 2, 2012 Report Share Posted November 2, 2012 Question :a) Ionized hydrogen atoms are accelerated from rest in the vacuum between two vertical parallel conducting plates . the potential differnce between the plates is V as a result of accelertation each ion gain energy of 1.9*10^-18 Jfind the value of V )b) the plates in (a) are replaced by a cell that has an emf of 12 V and internal resistance 5 ohms . A resistor R is connected in series with the cell . The energy transferred by the cell to an electron as it moves through the resistor is 1.44*10^-18 JShow that the value of R is 15.0 Ohms .I Know how to find a but how can it show the answer to b ??Voltage across R = (1.44*10ˆ-18)/(1.6*10ˆ-19) = 9VHence R = 9/((12-9)/5) = 15 ohms 1 Reply Link to post Share on other sites More sharing options...
eelnedross Posted November 2, 2012 Report Share Posted November 2, 2012 Why is the distance of 10pc in absolute magnitude taken? Because it could be 20, 30 or even 40pc! Can someone please explain spectroscopic parallax to me?I'm sorry for asking too much questions, my final Physics IB exam is in 1 week and a half. 1 Reply Link to post Share on other sites More sharing options...
Procrastination Posted November 2, 2012 Report Share Posted November 2, 2012 In spectroscopic parallax no use of parallax is made, kind of misleading. Used to measure distance from stars. It is a method that involves the luminosity and apparent brightness of a star. We can use analysis of spectra and Wien's displacement law to find the surface temperature of the star. Assuming the star is a main sequence star, we can use the H-R diagram to find its luminosity. Then if its apparent brightness is known, the distance to the star can be found using the relationship between apparent brightness and sitance:b= L/4d^2(pi)And yes you're right, but conveniently astronomers defined absolute magnitude of a start as the apparent magnitude it would have if it were 10 pc. Just because. It is just used to compare the stars Reply Link to post Share on other sites More sharing options...
eelnedross Posted November 10, 2012 Report Share Posted November 10, 2012 (edited) Can someone please adress the following syllabus points for me? I can't find them in my textbook and the exam is near!A.6.9 Outline qualitatively the action of liquid-crystal displays (LCDs).E.3.10 Explain how stellar distance may be determined using apparent brightness and luminosity.E.3.14 State the relationship between period and absolute magnitude for Cepheid variablesE.3.16 Determine the distance to a Cepheid variable using the luminosity–period relationship.Thank you very much for all of your support IBS! Edited November 10, 2012 by eelnedross Reply Link to post Share on other sites More sharing options...
MLI Posted November 22, 2012 Report Share Posted November 22, 2012 Can someone please explain what is Lenz's Law: "the direction of the induced e.m.f. is such that, if a complete circuit were available, the current directions would be oppose the change that caused the induced e.m.f."Thanks Reply Link to post Share on other sites More sharing options...
SerUmbras Posted November 25, 2012 Report Share Posted November 25, 2012 Okay, here goes. A change in emf is caused by a change in magnetic field, as this changes the current (think right hand rule).Think of a ring of wire. It has its own magnetic field (though it's negligible in this case). If the magnetic field in the centre of that ring was increased, by say putting a bar magnet into it, a change in emf will occur. This leads to a current being generated in a wire. What Lenz's law says is that the current that is generated will create a field that opposes the original change. In this case, it will try and oppose the increase, so it will be in the opposite direction. If the bar magnet was then pulled out of the ring, there would be a decrease in the magnetic field, so the induced current would be in the opposite direction.As for emf, that's just the force applied to the electrons to make a current ( I think, tell me if I'm wrong). So, the emf will be in the same direction as the current.Hope that helped. GL, HF, DD 1 Reply Link to post Share on other sites More sharing options...
The Rainbow Connection Posted January 4, 2013 Report Share Posted January 4, 2013 Hello everyone! I need a bit of help on this thermal physics question: A hot water vessel contains 3.0 dm3 at 45 °C. Calculate the rate that the water is losing thermal energy (in joules per second) if it cools to 38 °C over an 8.0 h period. Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted January 4, 2013 Report Share Posted January 4, 2013 Hello everyone! I need a bit of help on this thermal physics question: A hot water vessel contains 3.0 dm3 at 45 °C. Calculate the rate that the water is losing thermal energy (in joules per second) if it cools to 38 °C over an 8.0 h period.The specific heat capacity of water is 4200 J kgˆ-1 Kˆ-1 and 3.0 dm3 of water weighs 3 kg.So the thermal energy lost in cooling from 45 to 38 C = (45 - 38)*3*4200 = 88,200 J.Hence the average rate is 88,200/(8*60*60) = 3 J sˆ-1. 2 Reply Link to post Share on other sites More sharing options...
Fermat Posted January 20, 2013 Report Share Posted January 20, 2013 (edited) EDIT: Misread the question Edited January 23, 2013 by Fermat Reply Link to post Share on other sites More sharing options...
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