標題:
物理波動的問題-希望有心人幫幫我!
發問:
Q1一列波在水面上傳播。連續5個波峰的距離是120mm，而這列波在1.2s內的傳播距離是240mm。求波的波長、傳播速率和頻率，並以國際單位表示你的答案。答案:波長=0.03m 點解波長唔係0.024 請答解釋f=8除以1.2=6.67Hz 點解係波長係8?我最想知就係5個波峯，姐係幾多波長。請解釋希望盡可能最詳細解釋Q2一個聲波由音叉產生，並在空氣中傳播。波的波長是1.1m，而空氣質點在0.2秒內完成了60次振盪。問題-連續3個相鄰疏部的距離是多少?答案-2.2m 點解3個相鄰疏部係2.2m? 是否因為2個波長-2.2m... 顯示更多 Q1 一列波在水面上傳播。連續5個波峰的距離是120mm，而這列波在1.2s內的傳播距離是240mm。求波的波長、傳播速率和頻率，並以國際單位表示你的答案。 答案:波長=0.03m 點解波長唔係0.024 請答解釋 f=8除以1.2 =6.67Hz 點解係波長係8? 我最想知就係5個波峯，姐係幾多波長。請解釋 希望盡可能最詳細解釋 Q2 一個聲波由音叉產生，並在空氣中傳播。波的波長是1.1m，而空氣質點在0.2秒內完成了60次振盪。 問題-連續3個相鄰疏部的距離是多少? 答案-2.2m 點解3個相鄰疏部係2.2m? 是否因為2個波長-2.2m ，而2個波長有4個密部和3個疏部?若是，這個情況是否自然定律，或情況不同而導致? 希望有心者回應!
最佳解答:
Q1: If you draw out a digram, this will be clear to you. The distance between two succesive wave crest (波峰) is one wavelength (波長). Between three successive wave crest, there are two wavelengths. Hence, there are four wavelengths between 5 successive crest.Wavelength = 120/4 mm = 30 mm = 0.03 mSpeed of wave = 240/1.2 mm/s = 200 mm/s = 0.2 m/sFrequency = speed of wave/wavelength = 0.2/0.03 Hz = 6.67 Hz Q2: Distance between 3 successive rarefactions (疏部) = 2 x wavelength= 2 x 1.1 m = 2.2 mQ:而2個波長有4個密部和3個疏部? No. There are 3 rarefactions (疏部) and 2 compressions (密部) within two wavelengths. 2010-12-21 13:59:20 補充： Ans to the above question: This depends from which point of the wave you start counting. You coul dhave 3 compressions and 2 rarefactions within 2 wavelengths if you start the counting from a compression.
其他解答:
Q1 No. There are 3 rarefactions (疏部) and 2 compressions (密部) within two wavelengths. I want to know whether have 3 rarefactions and 2 compressions within two wavelengths in any cases?Does it is a natural case or there will be 2 compressions and 3rarefactions within two wavelengths in other case?
物理波動的問題-希望有心人幫幫我!
發問:
Q1一列波在水面上傳播。連續5個波峰的距離是120mm，而這列波在1.2s內的傳播距離是240mm。求波的波長、傳播速率和頻率，並以國際單位表示你的答案。答案:波長=0.03m 點解波長唔係0.024 請答解釋f=8除以1.2=6.67Hz 點解係波長係8?我最想知就係5個波峯，姐係幾多波長。請解釋希望盡可能最詳細解釋Q2一個聲波由音叉產生，並在空氣中傳播。波的波長是1.1m，而空氣質點在0.2秒內完成了60次振盪。問題-連續3個相鄰疏部的距離是多少?答案-2.2m 點解3個相鄰疏部係2.2m? 是否因為2個波長-2.2m... 顯示更多 Q1 一列波在水面上傳播。連續5個波峰的距離是120mm，而這列波在1.2s內的傳播距離是240mm。求波的波長、傳播速率和頻率，並以國際單位表示你的答案。 答案:波長=0.03m 點解波長唔係0.024 請答解釋 f=8除以1.2 =6.67Hz 點解係波長係8? 我最想知就係5個波峯，姐係幾多波長。請解釋 希望盡可能最詳細解釋 Q2 一個聲波由音叉產生，並在空氣中傳播。波的波長是1.1m，而空氣質點在0.2秒內完成了60次振盪。 問題-連續3個相鄰疏部的距離是多少? 答案-2.2m 點解3個相鄰疏部係2.2m? 是否因為2個波長-2.2m ，而2個波長有4個密部和3個疏部?若是，這個情況是否自然定律，或情況不同而導致? 希望有心者回應!
最佳解答:
Q1: If you draw out a digram, this will be clear to you. The distance between two succesive wave crest (波峰) is one wavelength (波長). Between three successive wave crest, there are two wavelengths. Hence, there are four wavelengths between 5 successive crest.Wavelength = 120/4 mm = 30 mm = 0.03 mSpeed of wave = 240/1.2 mm/s = 200 mm/s = 0.2 m/sFrequency = speed of wave/wavelength = 0.2/0.03 Hz = 6.67 Hz Q2: Distance between 3 successive rarefactions (疏部) = 2 x wavelength= 2 x 1.1 m = 2.2 mQ:而2個波長有4個密部和3個疏部? No. There are 3 rarefactions (疏部) and 2 compressions (密部) within two wavelengths. 2010-12-21 13:59:20 補充： Ans to the above question: This depends from which point of the wave you start counting. You coul dhave 3 compressions and 2 rarefactions within 2 wavelengths if you start the counting from a compression.
其他解答:
Q1 No. There are 3 rarefactions (疏部) and 2 compressions (密部) within two wavelengths. I want to know whether have 3 rarefactions and 2 compressions within two wavelengths in any cases?Does it is a natural case or there will be 2 compressions and 3rarefactions within two wavelengths in other case?
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