The transverse displacement of an harmonic wave on a stretched rope is y = 0.05
ID: 1507969 • Letter: T
Question
The transverse displacement of an harmonic wave on a stretched rope is y = 0.05 cos(2.9 x - 5.8 t), where x and y are in meters and t is in seconds.
1)
What is the amplitude of this wave?
A =
m
2)
What is the wavelength of this wave?
l =
m
3)
What is the speed with which this wave travels?
|v| =
m/s
4)
In what direction is this wave propagating?
+x
-x
+y
-y
+z
-z
none of the above
Your submissions:
A
Submitted:
Monday, April 25 at 7:23 PM
Feedback:
Right answer !
5)
What is the frequency (NOT the angular frequency) of this wave?
f =
Explanation / Answer
compare the given equation with
y = A*cos(k*x - w*t)
1) A = 0.05 m
2) k = 2.9
2*pi/lamda = 2.9
lamda = 2*pi/2.9
= 2.17 m
3) v = w/k
= 5.8/2.9
= 2 m/s
4) +x
5) f = w/(2*pi)
= 5.8/(2*pi)
= 0.923 Hz
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