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 = 5.3sin(0.026x +3.4t), wherex and y are expressed in centimeters andt is in seconds. Determine the following values. (a) the amplitude
Enter anumber. 1 cm
(b) the wavelength
Enter anumber. 2 cm
(c) the frequency
Enter anumber. 3 Hz
(d) the speed
Enter anumber. 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
Enter anumber. 6 cm/s
(g) the transverse displacement at x = 3.5 cm whent = 0.26 s
Enter anumber. 7 cm (a) the amplitude
Enter anumber. 1 cm
(b) the wavelength
Enter anumber. 2 cm
(c) the frequency
Enter anumber. 3 Hz
(d) the speed
Enter anumber. 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
Enter anumber. 6 cm/s
(g) the transverse displacement at x = 3.5 cm whent = 0.26 s
Enter anumber. 7 cm Enter anumber. Enter anumber. Enter anumber. Enter anumber. 5 +x -x    +y -y 5 +x -x    +y -y 5 Enter anumber. Enter anumber.

Explanation / Answer

Given : .               Equation of transverse wave is : .                             y= 5.3 sin(0.026x + 3.4t) . (a)   Amplitude is : A = 5.3 cm . (b)   We know that : .         Wave length() = 2 / k = 2 /0.026   = 76.92 cm . (c)    Freqency is : .         f  =   / 2   = 3.4 /2   = 1.7 Hz . (d)   Speed is : .         v = * f   = 76.92   * 1.7 = 130.76 cm /s . (e)   Maximum speed is : .       Vmax   = A /    = ----------- cm /s . (f)   We have : .          v = x / t   ==>   130.76 = (3.5 cm - x) / 0.26 .                            ==>   x   =   -----  cm . Solve the above . Hope this helps u!

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