A generator at one end of a very long string creates a wave given by y = (5.64 c

Question

A generator at one end of a very long string creates a wave given by y = (5.64 cm) cos[(/2)(2.06 m-1x + 6.44 s-1t)] and a generator at the other end creates the wave y = (5.64 cm) cos[(/2)(2.06 m-1x - 6.44 s-1t)] Calculate the (a) frequency, (b) wavelength, and (c) speed of each wave. For x 0, what is the location of the node having the (d) smallest, (e) second smallest, and (f) third smallest value of x? For x 0, what is the location of the antinode having the (g) smallest, (h) second smallest, and (i) third smallest value of x?

Explanation / Answer

y1 = (5.64 cm) cos[(1.03 x + 3.22 t)]

y2 = (5.64 cm) cos[(1.03 x - 3.22 t)]

w = 3.22 ===> f = w/2 = 1.61 Hz ............... (ans a)

k = 1.03 ======> Lambda = 2/k = 1.94 m ............(ans b)

v = lambda*f = 1.94*1.61 = 3.12 m/s .............(ans c)

Now y1 + y2 = 5.64*cos[(1.03 x + 3.22 t)] + 5.64*cos[(1.03 x - 3.22 t)]

   = 5.64*[cos[(1.03 x + 3.22 t)] + cos[(1.03 x - 3.22 t)]]

= 5.64[2*cos(1.03 x)*cos(3.22 t)] =  11.28*cos(1.03 x)*cos(3.22 t)

Now for nodes, cos(1.03 x) = 0 = cos(n + /2)

  (1.03 x) = (n + /2) =======> x = (0.97*n + 0.485) m

d) first smallest node, n = 0 =========> x = 0.485 m

e) Second smallest node, n = 1 =========> x = 1.455 m

f) Third smallest node, n = 2 =========> x = 2.425 m

Now for Antinodes, cos(1.03 x) = +-1 = cos(n)

  (1.03 x) = (n) =======> x = (0.97*n) m

d) first smallest Antinode, n = 0 =========> x = 0 m

e) Second smallest antinode, n = 1 =========> x = 0.97 m

f) Third smallest antinode, n = 2 =========> x = 1.94 m

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