Frequency in two different wires

In summary, the problem involves an aluminum wire and a steel wire joined together with a block of mass 10.0kg hanging from the end. Transverse waves are set up on the wire by a variable frequency source, with a node located at the pulley. The problem asks for the lowest frequency that will generate a standing wave with the joint as one of the nodes, and how many nodes are observed at this frequency. The solution involves finding the wave speeds in each section using the formula v = (F/(linear density))^0.5, and then determining the ratio of wavelengths in each section to find the frequency that produces an integral number of loops in both sections. The relationships between the number of loops and the wavelengths must be considered
  • #1
LandOfStandar
61
0
[SOLVED] Frequency in two different wires

Homework Statement



An aluminum wire. of length L1=60.0cm, cross-section area 1.00 x 10^-2 cm^2, and density 2.60g/cm^3, is joined to a steel wire, of density 7.80g/cm^3 and the same cross-section area. The compound wire, loaded with a block of mass m-10.0kg, is arranged so that the distance L2=86.6cm. Transverse waves are set up on the wire by an external source of variable frequency; a node is located at the pulley. (a) Find the lowest frequency that generates a standing wave having the joint as one of the nodes. (b) How many nodes are observed at this frequency?

Homework Equations

and

The Attempt at a Solution



f= v/(wavelength) = (n x v)/(2L)
v= (F/(linear density))

I know the F is 10.0kg x 9.80m/s^2 = 98.0N
...for the velocities do you do them saperate or together
...how do you get the wavelength
 
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  • #2
Day and hours have pasted and I have gotten not, just need to know if anyone have some suggestions on the approach to get me started.
 
  • #3
Please Help!
 
  • #4
I am lost with the question why is no one helping?
 
  • #5
The wave speed is different on each stretch of wire. Once you figure out the wave speeds, find the wavelengths using [itex]v = f \lambda[/itex].

(Look up the formula for wave speed; yours is not quite right.)
 
  • #6
I don't know the frequency or wavelength

v= (F/(linear density))^(.5)

v= ([10.0 x 9.8]/[2.60 x 1.00 x 10^-2])^(.5) = 61.4
v= ([10.0 x 9.8]/[7.80 x 1.00 x 10^-2])^(.5) = 35.4

I can get that but then I am lost
 
  • #7
LandOfStandar said:
I don't know the frequency or wavelength
The frequency is what you're asked to find. Since the frequency is constant over both segments of wire, what can you say about the ratio of the wavelength?

Then you just have to play around until you find a set of wavelengths that fit the conditions for a standing wave with a node at one end and at the joint. (Which end is attached to the source?)
 
  • #8
I don't know which end is attached to the source, what do I do with the nodes because part b ask for them?
 
  • #9
I am lost with what was previously said can anyone help
 
  • #10
Assume that you have nodes at each end and at the joint. Hint: You need to have an integral number of "loops" in each wire section to have a standing wave matching those conditions. How do the number of loops in each section compare? (Find their ratio.) Then figure out the smallest values that will give you an integral number of loops in both sections. Use that to find the frequency.
 
  • #11
Can't make sense

I have found how many nodes there are(I used the idea that L2 = 1.44 x L1, so maybe this is where I went wrong.) and how long the wavelength is. I am not sure I follow what is wrong with LandOfStandar's speed formula, but that may be my problem. Since I know the number of wavelengths, I should be able to use this to tell me frequency. I have different wave speeds in L1 and L2 which is what makes me confused.

my textbook says that v = wavelength x frequency

From the sound of Doc Al's posts this is incorrect?

If the two speeds are different then how can then this says to have the same frequency, the length of wavelengths must be different. If this is the case then I'm not sure how to relate the number of 'loops' that make up each length.
 
  • #12
ryukyu said:
my textbook says that v = wavelength x frequency

From the sound of Doc Al's posts this is incorrect?
No, that's perfectly correct.

If the two speeds are different then how can then this says to have the same frequency, the length of wavelengths must be different. If this is the case then I'm not sure how to relate the number of 'loops' that make up each length.
Yes, the wavelengths (and thus the size of the loops) are different in each section. How are they related?
 

Related to Frequency in two different wires

1. What is frequency in two different wires?

Frequency in two different wires refers to the rate at which an alternating current (AC) completes one full cycle in each of the two wires. It is measured in Hertz (Hz) and is typically the same in both wires if they are connected to the same power source.

2. How does the frequency in two different wires affect the flow of electricity?

The frequency in two different wires directly affects the flow of electricity, as it determines the speed at which the current alternates between positive and negative. A higher frequency means the current alternates more quickly, while a lower frequency means it alternates more slowly.

3. Can the frequency in two different wires be different?

Yes, the frequency in two different wires can be different if they are connected to different power sources. This can happen in situations where there are multiple power sources or if one wire is connected to an AC source and the other is connected to a DC source.

4. How can frequency in two different wires be measured?

Frequency in two different wires can be measured using a frequency meter or an oscilloscope. These tools measure the rate of change in the electric current and display it in Hertz (Hz).

5. What happens if the frequency in two different wires is not the same?

If the frequency in two different wires is not the same, it can cause issues with the flow of electricity and potentially damage electrical equipment. This is why it is important for wires connected to the same power source to have the same frequency.

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