What is the Length of an Aluminum Rod Creating a Standing Wave at 6180 Hz?

[SpringWater]

Homework Statement

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.

Homework Equations

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

 

[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.

 

[SpringWater]

When solving standing wave problems, its 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.

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 I've 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?

 

[SpringWater]

i forgot to include this picture

 

[TSny]

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

Note that you are considering the lowest frequency standing wave.

Also I've 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?

Are you sure that L/2 is the distance between a node and antinode?

 

[TSny]

i forgot to include this picture

The figure at the top left looks good.

 

[haruspex]

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.

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.

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

Because it's the lowest frequency at which resonance occurs.

Also I've 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?

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.