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) = 0.15sin(2.3x + 17.4t) The mass density of the rope is = 0.128 kg/m. x and y are measured in meters and t in seconds.

1)What is the amplitude of the wave?

2)What is the frequency of oscillation of the wave?

3)What is the wavelength of the wave?

4)What is the speed of the wave?

5)What is the tension in the rope?

6)At x = 3.4 m and t = 0.42 s, what is the velocity of the rope? (watch your sign)

7)At x = 3.4 m and t = 0.42 s, what is the acceleration of the rope? (watch your sign)

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

9)In what direction is the wave traveling?

+x direction

-x direction

+y direction

-y direction

+z direction

-z direction

10)On the same rope, how would increasing the wavelength of the wave change the period of oscillation?

a)period would increase

b)the period would decrease

c)the period would not change

Explanation / Answer

comparing with the general wave equation y(x,t) = A*sin(kx+wt)

here A is amplitude
k is wave number = (2*pi)/lamda

w is angular frequency = 2*pi*f...

y(x,t) = 0.15sin(2.3x + 17.4t)
1) A = 0.15m

2) f = 2*pi/w = 6.284/17.4 = 0.361 Hz...

3) lamda = 2*pi/k = 6.284/2.3 = 2.73 m

4) speed v = w/k = 17.4/2.3 = 7.56 m/sec...

5) T = v^2*mu = 7.56^2*0.128 = 7.315 N

6) v = dy/dt = 0.15*17.4*cos(2.3x + 17.4t)..
v = 0.15*17.4*cos((2.3*3.4) + (17.4*0.42)) = -2.18 m/sec..
7) a = dv/dt = -0.15*17.4*17.4*sin((2.3*3.4) + (17.4*0.42)) = -24.88 m/s^2
8) v= A*W/2 = (0.15*17.4)/2 = 1.305 m/sec...
9) -X direction
10) lamda = 2*pi/k...it is independent of timeperiod..so answer is

c)the period would not change

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