dkdqgav的部落格

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physics 3

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1.A piano tuner hears one beat every 3.5 s when trying to adjust two strings, one of which is sounding 440 Hz, so that they sound the same tone. How far off in frequency is the other string?2.A particular guitar string is supposed to vibrate at 199 Hz, but it is measured to actually vibrate at 211 Hz. By what... 顯示更多 1.A piano tuner hears one beat every 3.5 s when trying to adjust two strings, one of which is sounding 440 Hz, so that they sound the same tone. How far off in frequency is the other string? 2.A particular guitar string is supposed to vibrate at 199 Hz, but it is measured to actually vibrate at 211 Hz. By what percent should the tension in the string be changed to get the frequency to the correct value? (Your answer should be positive for an increase and negative for a decrease in the tension.)

最佳解答:

1. 另一條絃的頻率和440Hz每3.5秒便振動多了/少了一次，即每一秒差了 1/3.5 = 0.2857Hz(週)。 所以另一條絃的頻率可以是 440 + 0.2857 = 440.2857 Hz 或440 - 0.2857 = 439.7143 Hz 2. 由以下網址查出頻率和張力的關係 http://www.geocities.com/CapitolHill/Lobby/7049/strmpl.htm the fundamental relationship is: FREQ = sqrt(TENS/MPL)/LENG/2 Solving for the thing we need to know: MPL = 9800xTENS/FREQ/FREQ/LENG/LENG/4 where TENS is in kilograms (kg), FREQ is in cycles per second (cps), LENG is in meters (m), and MPL is in grams per meter (g/m). 即 頻率[f] 正比於 √(張力) (211/199) = √(T211/T199) (T211/T199) = 1.124 所以 絃的張力應 增加 12.4%

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