A transverse harmonic wave travels on a rope according to the following expressi

Question

A transverse harmonic wave travels on a rope according to the following expression:
y(x,t) = 0.18 sin(2.4x + 17.6t)
The mass density of the rope is 0.124 kg/m. x and y are measured in meters and t in seconds.
1) What is the amplitude of the wave? _________________ m
What is the frequency of oscillation of the wave? ________________ Hz
What is the wavelength of the wave? _________________ m
What is the speed of the wave? _______________ m/s
What is the tension in the rope? _________________ N
At x = 3.5m and t = 0.43 seconds , what is the velocity of the rope? _______________m/s
At x = 3.5m and t= 0.43 seconds , what is the acceleration of the rope? ______________ m/s^2
What is the average velocity of the rope during one complete oscillation of the rope? ________________ m/s
In what direction is the wave traveling?
On the same rope, how would increasing the wavelength of the wave change the period of oscillation?

Explanation / Answer

thus y=0.04*cos(6*(t -x/2)) = A*cos(w*(t –x/v); A=0.04m; - (a); v=2m/s; - (c); w=6 =2pi/T, hence period T =(pi/3) s; ? =v/T = (6/pi) m; - (b); +x; - (d); f=1/T = (3/pi) Hz; - (e); dy/dt =-A*w*sin(w*(t –x/2)) = = -0.04*6*sin(6*(3 –2/2)) = 0.129 m/s; - (f); d(dy/dt)/dt = -A*w^2 *cos(w*(t –x/2)) = = -0.04*6^2*cos(6*(3 –2/2)) = -1.215 m/s^2; - (g);

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