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From Giancoli\'s \"Physics for Scientists and Engineers.\" 15-20: A transverse t

ID: 1665949 • Letter: F

Question

From Giancoli's "Physics for Scientists and Engineers."
15-20: A transverse traveling wave on a cord is represented byD=0.48 sin (5.6x + 84t) where D and x are in meters and t inseconds. For this wave, determine a) the wavelength b) frequency c)velocity (magnitude and direction), d) amplitude, and e) maximumand minimum speeds of particles of the cord. I need to calculate velocity and frequency to find wavelenth,but I don't have the mass per unit length or tension value fortransverse wave. Amplitude is 0.48. Need help with minand max speeds, too. Thank you, Heather B. From Giancoli's "Physics for Scientists and Engineers."
15-20: A transverse traveling wave on a cord is represented byD=0.48 sin (5.6x + 84t) where D and x are in meters and t inseconds. For this wave, determine a) the wavelength b) frequency c)velocity (magnitude and direction), d) amplitude, and e) maximumand minimum speeds of particles of the cord. I need to calculate velocity and frequency to find wavelenth,but I don't have the mass per unit length or tension value fortransverse wave. Amplitude is 0.48. Need help with minand max speeds, too. Thank you, Heather B.

Explanation / Answer

(a) wavelength = 2 / k = 2 / 5.6   =    1.122 meters . (b) frequency = / 2 = 84 /2 =    13.37 Hz . (c) velocity = wavelength * freq = 1.122 * 13.37 =     15.0 m/s    in the negative x direction . (note: it travels in   -x direction because   5.6x and 84t   have thesame sign; if they had opposite signs, the wave would travel in the+x direction) . (d) amplitude = 0.48 meters  (given in equation) . (e) max speed = amplitude * angularfreq = 0.48 * 84 =    40.32m/s .      minimum speed is zero. . (f) max acc   = amplitude * ang freqsquared =   0.48 * 842=    3387m/s2      (in case you needthis)
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