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:
I imagine someone could do the math and figure out the range of possible notes that would come from a violin of that size.
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.
Hopefully they’ll take into consideration that the poor water bear never received any formal training and is not playing the instrument properly.
Kinda hard to play it backwards like that.
Jimi hendrix would have liked a word.