Physics - Error Measurement
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Physics - Error Measurement
2011-01-24 12:16 am
回答 (2)
2011-01-24 12:54 am
✔ 最佳答案
T = 2π√(m/k)k = m/[T/(2π)]^2
k = 4π^2m/T^2
if |ΔT/T| = 0.05 and |Δm/m| = 0.04
|Δk/k|
= |Δm/m| + 2|ΔT/T|
= 0.04 + 2(0.05)
= 0.14
The error of the measured k is 14% (D)
2011-01-24 12:58 am
Since k = (2.pi)^2.[m/T^2]
hence, dk/k = dm/m + 2(dT/T)
where dm/m and dT/T are the fractional errors of m and T respectively
but dm/m = 0.04 and dT/T = 0.05
thus, dk/k = 0.04 + 2 x(0.05) = 0.14
Therefore, the percentage errror of k is 14%
hence, dk/k = dm/m + 2(dT/T)
where dm/m and dT/T are the fractional errors of m and T respectively
but dm/m = 0.04 and dT/T = 0.05
thus, dk/k = 0.04 + 2 x(0.05) = 0.14
Therefore, the percentage errror of k is 14%
收錄日期: 2021-04-26 14:04:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110123000051KK00755