標題:
phy wave2
發問:
請到以下網址看問題 http://xcsdwe0.xanga.com/ 11(a) why it is a negative sine function? 59 請詳細解釋,thx
最佳解答:
11. The particle at x = 0 is going to start a +ve sinusoidal displacement. In other words, it has just finished going from the lowest position to the equilibrium position. Because in a displacement-distance graph, positions along the +ve x-axis indicate displacements in earlier times. Thus, there should be a trough to the right of the concerned particle at x = 0. That said, the wave-form is a negative sinusoidal curve. 59. Cross-sectional area of wire = 10^-2 cm^2 = 10^-6 m^2 Mass per unit length of aluminum wire = 2600 x 10^-6 kg/m = 2.6 x 10^-3 kg/m Mass per unit length of steel wire = 7800 x 10^-6 kg/m = 7.8x10^-3 kg/m Speed of wave in aluminium wire, Va = square-root[10g/2.6x10^-3] m/s Speed of wave in steel wire, Vs = square-root[10g/7.8x10^-3] m/s where g is the acceleration due to gravity (=9.81 m/s^2) Let f be the lowest possible frequency, hence, f = Va/(入a) = Vs/(入s) where 入a and 入s are the wavelengths at the aluminium and steel wires respectively 入a/入s = Va/Vs = square-root[10g/2.6x10^-3] /square-root[10g/7.8x10^-3] = square-root[7.8/2.6] = square-root(3) Let Na and Ns be the number of loops in the aluminium and steel wires respectively. Because each loop has a length of half a wavelength, thus (Na).(入a/2) = 0.6 and (Ns).(入s/2) = 0.866 Dividing, (Ns).(入s)/(Na).(入a) = 0.866/0.6 (Ns/Na).(入s/入a) = 0.866/0.6 Ns/Na = (0.866/0.6).(入a/入s) = (0.866/0.6).(square-root(3)) = 2.5 = 5/2 Therefore, at the lowest possible frequency, there are 5 loops on the steel wire and 2 loops on the aluminium wire. 入a = 2 x 0.6/2 m = 0.6 m 入s = 2 x 0.866/5 m = 0.3464 m Frequency = Va/入a = {square-root[10g/2.6x10^-3]}/0.6 Hz = 324 Hz Since there are (5+2) = 7 loops, the no. of nodes = 6
phy wave2
發問:
請到以下網址看問題 http://xcsdwe0.xanga.com/ 11(a) why it is a negative sine function? 59 請詳細解釋,thx
最佳解答:
11. The particle at x = 0 is going to start a +ve sinusoidal displacement. In other words, it has just finished going from the lowest position to the equilibrium position. Because in a displacement-distance graph, positions along the +ve x-axis indicate displacements in earlier times. Thus, there should be a trough to the right of the concerned particle at x = 0. That said, the wave-form is a negative sinusoidal curve. 59. Cross-sectional area of wire = 10^-2 cm^2 = 10^-6 m^2 Mass per unit length of aluminum wire = 2600 x 10^-6 kg/m = 2.6 x 10^-3 kg/m Mass per unit length of steel wire = 7800 x 10^-6 kg/m = 7.8x10^-3 kg/m Speed of wave in aluminium wire, Va = square-root[10g/2.6x10^-3] m/s Speed of wave in steel wire, Vs = square-root[10g/7.8x10^-3] m/s where g is the acceleration due to gravity (=9.81 m/s^2) Let f be the lowest possible frequency, hence, f = Va/(入a) = Vs/(入s) where 入a and 入s are the wavelengths at the aluminium and steel wires respectively 入a/入s = Va/Vs = square-root[10g/2.6x10^-3] /square-root[10g/7.8x10^-3] = square-root[7.8/2.6] = square-root(3) Let Na and Ns be the number of loops in the aluminium and steel wires respectively. Because each loop has a length of half a wavelength, thus (Na).(入a/2) = 0.6 and (Ns).(入s/2) = 0.866 Dividing, (Ns).(入s)/(Na).(入a) = 0.866/0.6 (Ns/Na).(入s/入a) = 0.866/0.6 Ns/Na = (0.866/0.6).(入a/入s) = (0.866/0.6).(square-root(3)) = 2.5 = 5/2 Therefore, at the lowest possible frequency, there are 5 loops on the steel wire and 2 loops on the aluminium wire. 入a = 2 x 0.6/2 m = 0.6 m 入s = 2 x 0.866/5 m = 0.3464 m Frequency = Va/入a = {square-root[10g/2.6x10^-3]}/0.6 Hz = 324 Hz Since there are (5+2) = 7 loops, the no. of nodes = 6
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