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 3200 kg/m3. The mass of the wire is small enough that its effect on the tension in the wire can be neglected. 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 43.0 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 30.0 Hz . What is the density of the liquid?

For anyone finding this in the future, Ryan has the correct answer.

I solved it! It is:

((m_rock*g)-(m_rock/(g*density_rock))*density_liquid*g)/(lower frequency squared) = (m_rock*g)/(higher frequency squared)

Then just solve for density_liquid :)

Close but it's actually,

((m_rock*g)-((m_rock*g)/(density_rock*density_liquid)))/(lower frequency squared) = (m_rock*g)/(higher frequency squared)

To solve this problem, we need to use the principle of wave speed in a string. The wave speed in a string can be calculated using the formula:

v = √(T/μ)

where v is the wave speed, T is the tension in the string, and μ is the linear mass density of the string.

In this case, we can find the linear mass density of the string using the formula:

μ = m/L

where m is the mass of the string and L is the length of the string.

Given that the length of the wire is 3.00 m and the mass of the wire is small enough to be neglected, the linear mass density of the wire (μ_w) is approximately 0 kg/m.

Now, let's consider the situation when the rock is in air. Using the formula for the wave speed, we can write:

v_air = √(T/μ_w)

Similarly, when the rock is submerged in the liquid, the wave speed can be expressed as:

v_liquid = √(T/μ_l)

where μ_l is the linear mass density of the wire when the rock is submerged.

We can equate the two wave speeds since the wire dimensions and tension remain the same:

√(T/μ_w) = √(T/μ_l)

Squaring both sides of the equation, we get:

T/μ_w = T/μ_l

Since μ_w is approximately 0 kg/m, we can simplify the equation to:

1/μ_l = 1/μ_w

Simplifying further, we find:

μ_l = μ_w

Therefore, the linear mass density of the wire is the same when the rock is in air and when it is submerged in the liquid.

Now, we know that the linear mass density of the rock (μ_r) can be calculated using the formula:

μ_r = m_r/L

where m_r is the mass of the rock and L is the length of the wire. Given the density of the rock (ρ_r = 3200 kg/m^3), we can find the mass of the rock:

m_r = ρ_r * V

where V is the volume of the rock. The volume of the rock can be calculated using its weight and density:

V = W/(ρ_r * g)

where W is the weight of the rock and g is the acceleration due to gravity.

Substituting the values, we can find the mass of the rock:

m_r = 164.0 N / (3200 kg/m^3 * 9.8 m/s^2)

Now, we can substitute the mass of the rock and the length of the wire in the equation to find the linear mass density of the wire:

μ_r = m_r / L

Finally, to find the density of the liquid (ρ_l), we can use the formula:

ρ_l = μ_r / A

where A is the cross-sectional area of the wire.

Unfortunately, the cross-sectional area of the wire is not given in the problem, so we cannot find the density of the liquid without this information.