# Homework Help: Standing Wave in a rod

1. Mar 3, 2013

### SpringWater

1. The problem statement, all variables and given/known data

An aluminum rod is clamped at the one- quarter position and set into longitudinal vibration by a variable-frequency driving source. The lowest frequency that produces resonance is 6180 Hz.
Find the length of the rod. The speed of sound in aluminum is 4860 m/s.

2. Relevant equations

3. The attempt at a solution

f= (V) / (λ) or f=((n)*(V)) / (λ) I am having a hard time understanding the concepts

I have tried to figure this out but I am at a loss, why if we clamp the rod at (L/4) does the equation become.. f=V/L

How does the position change the formula? And for f=n*V / λ if n=2 and λ=2L then it makes sense to me but from the given data how do I know there are two anti-nodes or second harmonics???

Any help in explaining how I can further understand this would be greatly appreciated.
Thank you

2. Mar 3, 2013

### TSny

When solving standing wave problems, it's a good idea to make a sketch showing the locations of all nodes and antinodes in the system. The diagram can then be used to find the relationship between λ and L.

Last edited: Mar 3, 2013
3. Mar 3, 2013

### SpringWater

So then...since the diagram is not given;

the clamp is the position of a node and given that the distance between a node and anti-node is =λ/4

so after creating a diagram w/o knowing how many nodes and anti-nodes are present in the system..i am assuming that since the rod is clamped at L/4 and the ends are esentially open so the anti-node would be at both ends, so a anti-node is at L/4 so therefore, there are two nodes.

BUT! how would I know that there are not three anti-nodes before reaching the L/4 length???

Also ive tried using distance between node and anti-node=λ/4 ---> =L/2 so then the adj-anti-node will be at the (L/2) mark and I can figure it out from there...is this correct?

Last edited: Mar 3, 2013
4. Mar 3, 2013

### SpringWater

i forgot to include this picture

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5. Mar 3, 2013

### TSny

Note that you are considering the lowest frequency standing wave.
Are you sure that L/2 is the distance between a node and antinode?

6. Mar 3, 2013

### TSny

The figure at the top left looks good.

7. Mar 3, 2013

### haruspex

Your first diagram is right. The second one has 3/4 λ on the left and 5/4 λ on the right, a ratio of 5:3 instead of 3:1.
Because it's the lowest frequency at which resonance occurs.
I'm not sure whether you're saying that was arrived at from your first diagram or from an alternative. You should have λ = L.