The density of copper is 9.8×10^3 and the density of gold is1.9×10^4. When the two wires of these two metals are held under the same tension the speed in the gold wire is found to be half of that of the copper. What is the ratio of the diameters of the two wires?

speed of what ?

To find the ratio of the diameters of the two wires, we can use the formula for the speed of a wave on a stretched string:

v = √(T/μ),

where v is the speed of the wave, T is the tension, and μ is the linear mass density (also known as density per unit length) of the wire.

Given that the speed in the gold wire is half that of the copper wire, we have:

v_gold = (1/2) * v_copper.

Since both wires are under the same tension, we can set their tension values equal to each other:

T_gold = T_copper.

Now, let's substitute these equations into the formula for the speed of the wave:

√(T_gold/μ_gold) = (1/2) * √(T_copper/μ_copper).

Since T_gold = T_copper, we can cancel out the tension terms, and we are left with:

√(μ_copper/μ_gold) = 1/2.

Squaring both sides of the equation gives us:

(μ_copper/μ_gold) = 1/4.

Dividing both sides of the equation by μ_copper gives us:

μ_gold/μ_copper = 1/(4μ_copper).

Now, let's substitute the given densities into the equation:

(1.9×10^4)/(9.8×10^3) = 1/(4μ_copper).

Simplifying this expression results in:

1.94 = 1/(4μ_copper).

To find the ratio of the diameters, we need to relate the linear mass density (μ) to the diameter (d) of the wire. The linear mass density is equal to the mass per unit length of the wire (m) divided by the cross-sectional area (A):

μ = m/A.

The mass (m) is related to the density (ρ) and the volume (V) as:

m = ρ * V.

The volume (V) of a cylindrical wire is given by:

V = π * r^2 * L,

where r is the radius and L is the length of the wire.

Substituting the expressions for mass and volume, we obtain:

μ = (ρ * V) / A,

μ = (ρ * (π * r^2 * L)) / (π * r^2),

μ = (ρ * L) / r^2.

Now, let's denote the linear mass density of the copper wire as μ_copper and the linear mass density of the gold wire as μ_gold.

μ_copper = (ρ_copper * L) / r_copper^2,
μ_gold = (ρ_gold * L) / r_gold^2.

Dividing these two equations gives us:

(μ_copper/μ_gold) = (ρ_copper / ρ_gold) * (r_gold^2 / r_copper^2).

Now, let's substitute the expression for the ratio of the densities into the equation:

1.94 = (ρ_copper / ρ_gold) * (r_gold^2 / r_copper^2).

Simplifying this expression gives us:

1.94 = (9.8×10^3 / 1.9×10^4) * (r_gold^2 / r_copper^2).

Converting the given densities to scientific notation:

1.94 = (0.51) * (r_gold^2 / r_copper^2).

Now, let's solve for the ratio of the diameters:

(r_gold^2 / r_copper^2) = 1.94 / 0.51,

(r_gold / r_copper)^2 = 3.8.

Taking the square root of both sides:

(r_gold / r_copper) = √3.8,

r_gold / r_copper = ±√3.8.

Since the ratio of the diameters must be positive, we take only the positive square root:

r_gold / r_copper = √3.8.

Therefore, the ratio of the diameters of the two wires is approximately equal to √3.8, which is approximately equal to 1.949.