The equation of a transverse wave traveling along a very longstring is given by

Question

The equation of a transverse wave traveling along a very longstring is given by y = 6.0sin(0.027x +5.0t), wherex and y are expressed in centimeters andt is in seconds. Determine the following values. (a) the amplitude
1 cm

(b) the wavelength
2 cm

(c) the frequency
3 Hz

(d) the speed
4 cm/s

(e) the direction of propagation of the wave 5 +x
-x
+y
-y


(f) the maximum transverse speed of a particle in the string
6 cm/s

(g) the transverse displacement at x = 3.5 cm whent = 0.26 s
7 cm (a) the amplitude
1 cm

(b) the wavelength
2 cm

(c) the frequency
3 Hz

(d) the speed
4 cm/s

(e) the direction of propagation of the wave 5 +x
-x
+y
-y


(f) the maximum transverse speed of a particle in the string
6 cm/s

(g) the transverse displacement at x = 3.5 cm whent = 0.26 s
7 cm 5 +x
-x
+y
-y
5 +x
-x
+y
-y
5

Explanation / Answer

This is in the form y=A sin(kx-t+) A=amplitude=6cm k=2/ 0.027=2/ =116.35cm =2f 5=2f f=2.5Hz speed v=dy/dt=6*5*-cos(0.027x+5t)cm/s The maximum speed is when cos=1 For displacement plug in the values. Hence we get by it.

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