Resonance of sound waves can be produced within an aluminum rod by holding the rod at its midpoint and stroking it with an alcohol-saturated paper towel. In this resonance mode, the middle of the rod is a node while the ends are antinodes; no other nodes or antinodes are present. What is the frequency of the resonance if the rod is 1.30 m long?

To determine the frequency of the resonance, we need to know the speed of sound in the aluminum rod. The speed of sound in a solid medium can be calculated using the formula:

v = f * λ

where:
v = speed of sound
f = frequency
λ = wavelength

In this case, since the rod is fixed at both ends, the length of the rod will determine the wavelength of the sound wave. The wavelength can be calculated using the formula:

λ = 2L / n

where:
L = length of the rod
n = number of nodes (in this case, n = 2)

Given that the length of the rod is 1.30 m and there are 2 nodes, we can substitute these values into the formula to calculate the wavelength.

λ = 2 * 1.30 m / 2
λ = 1.30 m

Now, we have the wavelength and we need to calculate the speed of sound in the aluminum rod.

Next, we need to know the formula to calculate the speed of sound in a solid. In general, the speed of sound in a solid medium depends on the material's elasticity and density. For a solid rod, the speed of sound can be approximated using the formula:

v = √(E / ρ)

where:
v = speed of sound
E = Young's modulus of elasticity
ρ = density

The exact values for Young's modulus and density may vary depending on the specific type of aluminum used, but we can use some approximate values to get an estimate.

For aluminum, the approximate values are:
- Young's modulus (E): 70 GPa
- Density (ρ): 2,700 kg/m^3

Substituting these values into the formula, we can calculate the speed of sound in the aluminum rod.

v = √(70 GPa / 2,700 kg/m^3)
v = √(70 * 10^9 N/m^2 / 2,700 kg/m^3)
v ≈ 6,418 m/s

Now we have the speed of sound and the wavelength. We can use the formula v = f * λ to calculate the frequency.

6,418 m/s = f * 1.30 m

Solving for f:

f = 6,418 m/s / 1.30 m
f ≈ 4,940 Hz

Therefore, the frequency of the resonance in the aluminum rod is approximately 4,940 Hz.

I never found it necessary to soak the towels.

the rod is 1/2 wavelength, so wavelength=2.6m