A vibrating stretched string has length 139 cm, mass 38 grams and is under a tension of 38 newtons.

What is the frequency of its 5th overtone?
f_{6} =

If the source of the wave is an open organ pipe of the same length as the wire and the temperature of the air is 20^oC, what would be the frequency of the 5th harmonic?
f_{5} =

To find the frequency of the 5th overtone (f₆) of a vibrating stretched string, we need to use the formula for the fundamental frequency (f₁) of a vibrating string:

f₁ = (1/2L) √(T/μ)

Where:
L = length of the string (in meters)
T = tension in the string (in newtons)
μ = linear mass density of the string (in kg/m)

Given:
Length (L) = 139 cm = 1.39 m
Mass (m) = 38 grams = 0.038 kg
Tension (T) = 38 newtons

We need to find the linear mass density (μ) which is given by the formula:

μ = m/L

Substituting the values, we get:

μ = 0.038 kg / 1.39 m = 0.0273 kg/m

Now, substitute the values of L, T, and μ into the formula for the fundamental frequency (f₁) to find the fundamental frequency:

f₁ = (1/2(1.39)) √(38/0.0273)
f₁ = 0.359 Hz

The frequency of the 5th overtone (f₆) is given by the formula:

f₆ = 6f₁

Substituting the value of f₁, we get:

f₆ = 6(0.359)
f₆ = 2.154 Hz

Therefore, the frequency of the 5th overtone is 2.154 Hz.

Now, let's move on to finding the frequency of the 5th harmonic in an open organ pipe.

In an open organ pipe, the fundamental frequency (f₁) is given by:

f₁ = v/2L

Where:
L = length of the pipe (in meters)
v = speed of sound in air (in m/s)

In this case, we are told that the length of the open organ pipe is the same as the length of the wire, which is 1.39 m.

To find the speed of sound in air (v), we use the formula:

v = 331.5 * √(1 + T/273)

Where:
T = temperature of the air (in degrees Celsius)

Given:
T = 20°C

Substituting the value of T, we get:

v = 331.5 * √(1 + 38/273)
v ≈ 331.5 m/s

Now, substituting the values of L and v into the formula for the fundamental frequency (f₁), we get:

f₁ = 331.5/2(1.39)
f₁ ≈ 119.071 Hz

The frequency of the 5th harmonic (f₅) is given by the formula:

f₅ = 5f₁

Substituting the value of f₁, we get:

f₅ = 5(119.071)
f₅ ≈ 595.354 Hz

Therefore, the frequency of the 5th harmonic in an open organ pipe is approximately 595.354 Hz.