WAVE MC 1 題 again

2012-11-09 10:35 pm

a uniform wire is clamped at both ends which are 0.4m apart. the tension in the wire is 65N and the area of its cross-section is 4.5x10^-7 m^2 . when the wire is plucked to produce transverse vibrations, the frequency of the fundamental note obtained is 160Hz . the density of the wire is?

ans: 8800kg m^-3

thanks!

回答 (1)

天同

2012-11-10 12:44 am

✔ 最佳答案

Since the wire is vibrating at fundamental note, wavelength of wave 入= 2 x length of wire. Hence, 入 = 2 x 0.4 m = 0.8 m

velocity of wave v = frequency x wavelength
v = 160 x 0.8 m/s = 128 m/s

But v = square-root[T/d]
where T is the wire tension and d is the linear density (mass per unit length) of the wire.

thus, v^2 = T/d
128^2 = 65/d
d = 3.967 x 10^-3 kg/m

Since density p = d/A
where A is the cross-sectional area of the wire
thus, p = 3.967 x 10^-3/4.5 x10^-7 kg/m^2 = 8800 kg/m^3

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