Minimum frequency for a point to have maximum amplitude in standing wave

[Eitan Levy]

Homework Statement

We are given a rope with a mass of 20g and length of 1.5m.
The tension of the rope is 5N, and we are holding it in the edges.
Find the minimal frequency so that a point with a distance of 0.3m from one edge will have maximum amplitude.

Relevant Equations

f=vn/2L

When I tried using the equations the only thing I could see is that it is impossible for such point to be an anti-node. In this case, how do I find the frequency? The answer is not even with the form of v*n/2L which is very confusing to me, I thought that the frequency of a standing wave must have that form?

The answer is 30Hz.

Can someone please explain how to get this? How does that even make sense?

 

[BvU]

Hi,

You don't explain what v is
You miss a relevant equation that features the tension
Idem the mass (or the mass per unit length)
And perhaps a sketch is also extremely useful

Fix all that and then you can perhaps discover a path to the solution ...

 

[Eitan Levy]

We have the tension and the density of the mass. Sqrt(5*75) is supposed to be equal to v. 2L is 3.
The answer doesn't fit the formula of the frequency.

 

[BvU]

Is that 75 inches or 75 furlongs ?
Ah, m/kg ! so$$v=\sqrt {T\over \mu}\approx 19.4 \text{ m/s}$$
is the relevant equation I missed in your post.

 

[Eitan Levy]

Is that 75 inches or 75 furlongs ?

75 is the length divided by the mass.

 

[BvU]

impossible for such point to be an anti-node

I start to see your problem. You get resonances at multiples of 6.5 Hz and never an antinode at L/5, because they occur at L/2n

Conclusion: error in the problem statement