A transverse harmonic wave travels on a rope according to the following expressi
ID: 1296012 • Letter: A
Question
A transverse harmonic wave travels on a rope according to the following expression:
y(x,t) = 0.17sin(2.5x + 17.2t)
The mass density of the rope is ? = 0.103 kg/m. x and y are measured in meters and t in seconds.
1)
What is the amplitude of the wave?
m
Your submissions:
2)
What is the frequency of oscillation of the wave?
Hz
Your submissions:
3)
What is the wavelength of the wave?
m
Your submissions:
4)
What is the speed of the wave?
m/s
Your submissions:
5)
What is the tension in the rope?
N
Your submissions:
6)
At x = 3.4 m and t = 0.47 s, what is the velocity of the rope? (watch your sign)
m/s
Your submissions:
7)
At x = 3.4 m and t = 0.47 s, what is the acceleration of the rope? (watch your sign)
m/s2
Your submissions:
8)
What is the average speed of the rope during one complete oscillation of the rope?
m/s
Your submissions:
9)
In what direction is the wave traveling?
+x direction
-x direction
+y direction
-y direction
+z direction
-z direction
Your submissions:
10)
On the same rope, how would increasing the wavelength of the wave change the period of oscillation?
the period would increase
the period would decrease
the period would not change
Explanation / Answer
a) A = 0.17
2) w = 17.2
so f = 17.2/(2*pi)= 2.74 Hz
3)
k = 2 pi/lambda = 2.5
lambda = 2*pi/2.5= 2.51 m
4) v = lambda f = 2.74*2.51= 6.88 m/s
5) v = sqrt(T/mu)
6.88 = sqrt(T/0.103)
T= 4.88 N
6) v = dx/dt = 0.17*17.2*cos(2.5*x + 17.2 t)
= 0.17*17.2*cos(2.5*3.4 + 17.2*0.47)
=-1.87 m/s
7) a = dv/dt = -0.17*17.2^2*sin(2.5*3.4 + 17.2*0.47) = 38.6 m/s^2
8) average speed is 0
9) kx + wt so -x direction
10)lambda f = constant
so increasing lambda would decrease f which would mean T increases
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