• glorkon@lemmy.world
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    1 day ago

    Well well well.

    Suppose a normal violin has strings of ~33cm length. The E-string would have a frequency of ~660 Hz. Let’s shrink that down to tardigrade dimensions (according to Google, it’s about 400μm).

    I’m just going to assume the tardigrade violin has a string length of 60μm.

    The frequency of strings also depends on tensile stress and mass density - let’s just assume that these scale proportionally.

    So we can use the formula: f∝1/L (basically means, half the size means double the frequency).

    Let’s calculate the scale factor s for the frequency:

    L(real) = 330mm, f(real) = 660 Hz L(tardi) = 60μm.

    s = L(real) / L(tardi) = 0.33 / 6 * 10⁻⁵ = 5500.

    This means that the frequency of the tardigrade E-string would be:

    f(tardi) = f(real) * s = f(real) * 5500 = 660Hz * 5500 = 3,630,000Hz = 3.63 megahertz, which is 181.5 times above than the human limit of 20kHz.

    Difference in octaves… log2(3.63 Mhz / 660 Hz) = 15.7

    That means the tardigade E-string is almost 16 octaves above the human one.