Ondas estacinarias ejemplos

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ONDAS ESTACIONARIAS Y RESONANCIA
89. (a) L =

l1 , 2

F F

l1 = 2L = 2(3.0 m) = 6.0 m .
L 3.0 m l3 = 1.5 = 1.5 = 2.0 m .

(b) L = 1.5l3,

90.

(a) f2 = 2f1 =2(100 Hz) = 200 Hz . (b) f3 = 3f1 = 3(100 Hz) = 300 Hz .

91.

f3 = 3f1,

F

f1 =

92.

(a) f1 =

12 m/s v = = 1.5 Hz. So the answer is yes , 15 Hz is the 10thharmonic. 2L 2(4.0 m) 250 m/s v = = 62.5 Hz. 2L 2(2.0 m)

f3 450 Hz = = 150 Hz . 3 3

(b) No , 20 Hz is not a harmonic. 250 m/s v = = 312.5 Hz. 0.80 m l f 312.5 Hz = =5. fo 62.5 Hz

93.

f=

f1 =

So n =

95.

v=

fn =

nv 40 m/s = n = 10n Hz. 2L 2(2.0 m)

FT = m

40 N = 40 m/s. 2.5 ´ 10-2 kg/m

So the frequencies ofthe first four harmonics are 10 Hz‚ 20 Hz‚ 30 Hz‚ and 40 Hz .

97.

(a) The length of the string should be shortened because a shorter string has a shorter wavelength,therefore a higher frequency given the speed is a constant. (b) There is a node where the finger is placed. So the longest possible wavelength is f= 4v 8 v 8 8 v v = = = =f1 = (440 Hz) = 503 Hz . 7L 7 2L 7 7 l 7L/4 7 l = L. 2 8

101.

(a) The distance between two successive nodes is

l . So l = 2(6.0 cm) = 12 cm . 2

(b) Theanti-nodes are halfway between the nodes. So they are at 3.0 cm‚ 9.0 cm‚ 15 cm . 104.

v=

105.

332 m/s v l1 = f = 276 Hz = 1.2 m , 1 f1 = So f2 =

332 m/s v = = 2.8 ´ 102Hz . 2L 2(0.60 m) or

FT = m

550 N = 332 m/s. (3.0 ´ 10-3 kg)/(0.60 m)

l1 = 2L = 1.2 m.

The 4th harmonic and the 2nd harmonic vibrate in 4 and 2 loops,respectively. f4 420 Hz = = 210 Hz . 2 2

106.

fn = So

n 2L

FT , m

F

FT¢ ( f1¢ )2 (440 Hz)2 = = 0.956. 2 = FT (f1) (450 Hz)2

FT¢ = 0.956(500 N) = 478 N .

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