Set 6

(a) 8.49 m/sec
(b) 0.424n hz [where n is a whole number]

To solve this problem we will use the formula v_w = SQRT [ Tension / (mass/length) ].

The linear mass density, or mass per unit length (mass/length), of the tubing is given as 0.125 kg/m.  The tension in the tubing is given as 9 newtons, 9 N.

v_w = SQRT [ 9 / (0.125) ] = SQRT (72) = 8.49 m/sec

To determine the natural frequencies of the tubing, we must first determine its fundamental frequency. All natual resonance frequencies will be multiples of this value.

The fundamental frequency will have one loop in its waveform.  That loop equals 1/2 the fundamental wavelength.  If the tubing is 10 meters long, then 1/2 l = 10  ® l = 20 meters.

Using v_w = fl ® f =8.49 / 20 = 0.424 hz

To solve this problem we will use the formula v_w = SQRT [ Tension / (mass/length) ].

The linear mass density, or mass per unit length (mass/length), of the string is given as 2.5 x 10^-2 kg/m.  The tension in the string is given as 40 newtons, 40 N.

v_w = SQRT [ 40 / (0.025) ] = SQRT (1600) = 40.0 m/sec

To determine the natural frequencies of the string, we must first determine its fundamental frequency. All natural resonance frequencies will be multiples of this value.

The fundamental frequency will have one loop in its waveform.  That loop equals 1/2 the fundamental wavelength.  If the string is 2.0 meters long, then 1/2 l = 2.0  ® l = 4.0 meters.

Using v_w = fl ® f = 40.0 / 4.0  = 10.0 hz