The equation of a transverse wave traveling along a very long string is given by
ID: 1457182 • Letter: T
Question
The equation of a transverse wave traveling along a very long string is given by y = 6.4 sin(0.028x + 4.5t), where x and y are expressed in centimeters and t is in seconds. Determine the following values.
(a) the amplitude Correct: 6.4 cm
(b) the wavelength Correct: 71.43 cm
(c) the frequency: Hz
(d) the speed: 160.71 cm/s
(e) the direction of propagation of the wave +x -x +y or -y: -x
(f) the maximum transverse speed of a particle in the string: cm/s
(g) the transverse displacement at x = 3.5 cm when t = 0.26 s: cm
Bold parts need answers with explanations please!
Explanation / Answer
Compare the given equation with standrad equation
y = A*sin(kx + w*t)
so, A = 6.4 cm
k = 0.028*pi
w = 4.5*pi
C) f = w/(2*pi)
= 4.5*pi/(2*pi)
= 2.25 Hz
f) maximum transevrse speed, vmax = A*w
= 6.4*4.5*pi
= 90.5 cm /s or 0.905 m/s
g) at x = 3.5 cm, t = 0.26
y = 6.4*sin(0.028*pi*3.5 + 4.5*pi*0.26)
= 6.4*sin(3.983) (here use calculator in radians mode)
= -4.77 cm
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