Question

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

y(x,t) = 0.14sin(2.3x + 17.8t)
The mass density of the rope is μ = 0.107 kg/m. x and y are measured in meters and t in seconds.
I'm missing 6, 7 and 8

1)

What is the amplitude of the wave? 0.14m 

2)

What is the frequency of oscillation of the wave? 2.8334Hz 

3)

What is the wavelength of the wave? 2.73m 

4)

What is the speed of the wave? 7.74m/s

5)

What is the tension in the rope? 6.41N 

6)

At x = 3.5 m and t = 0.4 s, what is the velocity of the rope? (watch your sign)_____ m/s 

7)

At x = 3.5 m and t = 0.4 s, what is the acceleration of the rope? (watch your sign) ______m/s² 

8)

What is the average speed of the rope during one complete oscillation of the rope? 

Answer 1

Given, y (a, t) = 0.14 Sin (2-3x +17.8t) to General form, y (at) = A Sim (ka + wt). A Sin (21 x + 20 ft) 6 velocity of rope, vse = 0.14 x 17.8 Cos (232 +17.84) => V = 2.492 Cos (2.3 2 + 17.8 ) At, n= 3.5m, t= 0.45 N = 2.492 Cos [(23 x 3.5) + (17-8 x 0-477 V = 2:40516 m/s) l acceleration of rope, az [2.492 Cos(2.8 2 + 17-24)

=> 2=-2.492x17.8 Sin (2.3 2 +17.8t) >> a = - 44.3576 Sim (2.32 +17,8t) At n=3.5m, t = 0.45 >> a = - 44:8576 sra[ (2.843.5) + (17-8X0-9) -> a = -1.6076 m/ Distances travel = 4x amplitude d = 4x 0.14 m de 0.56 m one complete oscillation T = 21 = 2 2 17.8 See Average speed = 2 - my 0.56 X 17.8 - 2x3.14 = 1.587 m/
with regards. Thanks you.