A uniform 80 cm guitar string is fixed at both ends at A and B. A fingertip is touching the string at C and plucked between A and C to produce transverse vibrations, the frequency of the fundamental note is obtained is 880 Hz. The total mass of the string is 1.5g. The distance between A and C is 20 cm.
What is the tension of the string?
Ans: 232N
求解~ 謝謝!
Physics DSE Wave MC
2014-03-22 6:53 am
回答 (1)
2014-03-22 7:29 am
✔ 最佳答案
Since AC = 20 cm = 0.2 mwavelength = 2 x 0.2 m = 0.4 m
Speed of wave on string = 880 x 0.4 m/s = 352 m/s
Mass per unit length of string = (1.5/1000)/0.8 kg/m = 1.875 x 10^-3 kg/m
Hence, 352^2 = T/(1.875 x 10^-3)
where T is the string tension
T = (1.875 x 10^-3) x 352^2 N = 232 N
收錄日期: 2021-04-21 22:39:51
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