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A string with a fixed frequency vibrator at one end forms a standing wave with 4 antinodes when under tension T1. When the tension is slowly increased, the standing wave disappears until tension T2 is reached with no resonances occurring between the two tensions. How many antinodes are there in this new standing wave?

There would be 8 antinodes, right? Because as you jump up to the next frequency, the number of antinodes doubles.

Actually, no, it would be 5 antinodes, right?

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closed as too localized by Manishearth Jun 17 '13 at 9:12

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The relevant equation to use for a standing wave string under some tension is: $$ f_n = \frac{n}{2 L} \sqrt{\frac{F_T}{\rho}}$$ where $\rho$ is the mass density of the string and $F_T$ is the tension force.

For a fixed frequency we now have, after rearranging the above equation:

$$ F_{T,n} = \rho \left(\frac{2 L f_{\mathrm{fixed}}}{n}\right)^2 $$

When you increase the tension until you find a new standing wave pattern you must satisfy the above equation.

Remember that the letter n = 1 for the first standing wave pattern. If you have 4 antinodes then you have n = 4 in the above equation. The next standing wave pattern would be n = 5.

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