A large rock that weighs 164.0 N is suspended from the lower end of a thin wire that is 3.00 m long. The density of the rock is 3200kg/m3. The mass of the wire is small enough that its effect on the tension in the wire can be ignored. The upper end of the wire is held fixed. When the rock is in air, the fundamental frequency for transverse standing waves on the wire is 42.0 Hz. When the rock is totally submerged in a liquid, with the top of the rock just below the surface, the fundamental frequency for the wire is 28.0 Hz. What is the density of the liquid?

  ρ_liquid = 1,778 10³ kg/m³

Explanation:

In a wire the speed of the wave produced is

          v = √T/μ

wave speed is also related to frequency and wavelength

           v = λ f

as they indicate that the frequency is the fundamental, there must be a single antinode in the center

         L = 2λ

         λ = 2L

Using the translational equilibrium equation

         T -W = 0

         T = W

let's substitute

         2L f = √ W /μ

         μ = W / 4 L² f²

         μ = 164 / (4 3² 42²)

         μ = 2.5825 10⁻³ kg / m

this is the linear density of the wire

Now let's analyze when the rock is submerged in a liquid, the thrust acts on the rock

             B = ρ g V

when writing the equilibrium equation we have

              T + B -W = 0

              T = W - B

               T = W - ρ_liquid g V                    (1)

to find the volume of the rock we use the concept of density

              ρ_body = M / V

              V = M /ρ_body

              W = M g

              V = W / g ρ_body

              V = 164 / (9.8 3200)

              V = 0.00523 m³

               V = 5.23 10⁻³ m³

The speed of a wave on a string is

              v = √ T /μ

speed is also related to wavelength and frequency

              v = λ f

indicate that the frequency formed is the fundamental one, therefore it has a single antinode in the center and two nodes at the fixed points

             L = λ/ 2

             λ= 2L

we substitute and look for tension

             2L f = √T /μ

              T = 4L² f² μ

              T = 4 3² 28² 2.5825 10⁻³

              T = 72.888 N

We already have the volume of the rock and the tension on the rope, we can substitute in equation 1 and find the density of the liquid

               T= W – ρ_liquid g V  

               ρ_liquid = (W -T) / gV

               ρ_liquid = ( 164 – 72,888) / ( 9,8 5,23 10⁻³)

               ρ_liquid = 1,778 10³ kg/m³