HW 11a help & hint
#1 A hand holding a rope moves up and down to create a transverse wave on the rope.
The hand completes an oscillation in 1.9 s, and the wave travels along the string at 1 m/s.
The amplitude of the wave is 0.05 m.
a.) Find the frequency at which the crests pass a given point in space.
b.) Find the distance between two adjacent crests on the wave.
c.) There is a blue spot drawn onto the rope with a magic marker. Find the distance this
spot travels in one period.
d.) If the mass per unit length of the string is 8•10-4 kg/m, what is the tension in the string?
Someone asked:
"I am having trouble for 1c and d
c)isnt it just distance = vT, which would be 1m/s times 1.9s?
d)isnt it F= v^2/mass per unit of length?
Im not sure what im missing here..."
My answers:
Recall that we talked during the class. when wave travels, the media stay where they are.
When we play the "people wave", each person stay at their seat and just oscillate up and
down following the front person, so that the "wave" is spread out but each of us are still
there.
So the media doesn't move as the speed of wave. The media particles only oscillate with
amplitude A around their equilibrium center.
In each period of wave, object oscillating will cover a distance of 4A, but not go all the
way to distance of vT.
v=sqrt(F/mass per-unit-length), do math a little more carefully, you find that
v^2=F/(mass per unit length)
and F=_____________________ = ______
ALso pay attention to unit. Convert them to SI unit first.
Someone asked: Does this mean that distance is only related to amplitude?
Man, Weining:
a transverse wave:
For example, wave propagates in horizontal direction, but the string (spots on string)
oscillate in vertical direction.
Wavelength in horizontal direction.
Amplitude in vertical direction.
b.) Find the distance between two adjacent crests on the wave.
This distance is wavelength in the wave moving direction. Not related to amplitude.
c.) There is a blue spot drawn onto the rope with a magic marker. Find the distance this
spot travels in one period.
This distance is the distance of oscillation (vertical) for a single spot on the string. This is
only related to amplitude.
#2 A nylon guitar string has a linear density of 7.2g/m and is under a tension of 146N.
The fixed supports are 90cm apart. The string is oscillating in the standing wave pattern
shown below.
a.)Speed
student: I went to a help session and we used
v=sqrt(F/(m/L))
F=146N and m/L=7.2g/m. i used that equation with those numbers and it was wrong. can
you please tell me what i am doing wrong?
Man, Weining:
The key point is that you need to work on right units. convert g to kg, and cm to m
you have to change 7 g/m to kg/m
7 g/m is how many kg/m?
Only if you convert to SI unit: kg and meter,
you will get speed in SI unit m/s
waves travel pretty fast in thin strings like that.
Student: For part B it asks you to find the wavelength in cm. I thought wavelength was 2
times L. 2*90cm= 180 cm is wrong.
how am i supposed to find the wavelength?
Man, Weining: look at the pattern shown in the picture.
Recall what you saw in the class.
2 times L is the first harmonic wavelength.
In that case from the left end to the fight end, it covers half wavelength and there are no
nodes in between.
And there can be many other harmonic wavelengths for the same string. wavelength =
2L/N,
N is a integer.
They ask you to find the wavelength of the wave pattern shown in the picture.
By looking the picture, you know how long is the string, (between both ends), you can
also see from the left end point to which point you covers a wavelength. You can see that
your wavelength is how much portion of L.....
same thing for the last question.
By looking at the picture, you can see that half wavelength is what portion of L.
And the total wavelength is more than L, which is equal to ______???
Attention that for pipe with one open end, wavelength = 4L/N. See the pictures on
textbook and on lecture notes.
#3 Attention that in question 3,
For pipes with one open end, wavelength = 4L / N ,where N are odd numbers
For question3a, you can read the figure to find out one L is how much wavelength and then find
out wavelength. Then frequency. For question 3b, you can look at your text book or my lecture
note to find out that for pipes with one closed end, the FUNDAMENTAL wavelength will be how
many times of pipe length. Notice that the close end must have a node, and the open end must
be crest (maximum).
hw 11 #5 (b),
It says that "recall the third harmonic in this case is standing wave with the second lowest
frequency) Why is "the third harmonic in this case is standing wave with the second
lowest frequency"?
My answer:
for pipes with one closed end, wavelength = 4L/ N, here N is a odd number
lowest wavelength = 4L
second lowest wavelength = 4L/3
It is called "the third harmonic", because it is the fundamental wavelength / by 3.
There is no second harmonic (4L/2) for this pipe.
It's just how it is called.
The Nth harmonic is respect to the fundamental wavelength.