Frequency of oscillation for a stretched string:

$\overline{){\mathbf{f}}{\mathbf{=}}\frac{\mathbf{n}}{\mathbf{2}\mathbf{L}}{\mathbf{\xb7}}\sqrt{\frac{\mathbf{T}}{\mathbf{\mu}}}}$

Solving for n:

$\overline{){\mathbf{n}}{\mathbf{=}}{\mathbf{2}}{\mathbf{L}}{\mathbf{f}}{\mathbf{\xb7}}\sqrt{\frac{\mathbf{\mu}}{\mathbf{T}}}}$

The initial nodes, n_{i} = 2, and initial tension, T_{i}.

The final tension is 2T_{i}.

We'll solve for new nodes, n_{final}, as follows:

${\mathit{n}}_{\mathbf{i}}\mathbf{=}\mathbf{2}\mathit{L}{\mathit{f}}_{\mathbf{1}}\mathbf{\xb7}\sqrt{\frac{\mathbf{\mu}}{{\mathbf{T}}_{\mathbf{i}}}}$

${\mathit{n}}_{\mathbf{f}}\mathbf{=}\mathbf{2}\mathit{L}{\mathit{f}}_{\mathbf{1}}\mathbf{\xb7}\sqrt{\frac{\mathbf{\mu}}{\mathbf{2}{\mathbf{T}}_{\mathbf{i}}}}$

The figure shows a standing wave on a string.

Suppose the tension is doubled while the frequency shaking the string is held constant. Will there be a standing wave? If so how many antinodes will it have? If not, why not?

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