A transverse harmonic wave travels on a rope according to the following expressi
ID: 1305539 • Letter: A
Question
A transverse harmonic wave travels on a rope according to the following expression: y(x,t) = 0.14sin(2.5x + 17t) The mass density of the rope is ? = 0.113 kg/m. x and y are measured in meters and t in seconds.
You must use my numbers to receive the points.
1)
What is the amplitude of the wave?
m
2)
What is the frequency of oscillation of the wave?
Hz
3)
What is the wavelength of the wave?
m
4)
What is the speed of the wave?
m/s
5)
What is the tension in the rope?
N
6)
At x = 3.8 m and t = 0.43 s, what is the velocity of the rope? (watch your sign)
m/s
7)
At x = 3.8 m and t = 0.43 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?
m/s
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?
the period would increase
the period would decrease
the period would not change
Explanation / Answer
compare the given equation with
y(x,t) = A*sin(kx + w*t)
A = 0.14 m
w = 17 rad
k = 2.5 m^-1
1) A = 0.14 m
2) f = w/2*pi = 17/2*pi = 2.707 Hz
3)
k = 2*pi/lamda
==> lamda = 2*pi/k
= 2*pi/(2.5)
= 2.512 m
4) v = w/k
= 17/(2.5)
= 6.8 m/s
5) v = sqrt(T/mue)
==> T = v^2*mue
= 6.8^2*0.113
= 5.225 N
6) y = 0.14*sin(2.5*x + 17*t)
= 6.5 cm
v = dy/dt
= 0.14*17*sin(2.5*x+17*t)
at x = 3.8, t = 0.43 s
v = 0.14*17*cos(2.5*3.8+17*0.43)
= -0.063 m/s
7) a = dv/dt
= -0.14*17^2*sin(2.5*x+17*t)
at x = 3.9, t = 0.4 s
a = -0.14*17^2*sin(2.5*3.8+17*0.43)
= -0.12 m/s^2
8) T = 1/f = 1/2.707 = 0.3694 s
Vavg = 4*A/T = 4*0.14/0.3694 = 1.516 m/s
9) -x direction
10) the period would increase
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