Question:

A string of length 50 cm and a tension of 80 N is fixed at both ends. The string is set into oscillation, producing standing waves. When a single standing wave is formed on the string, there are 8 nodes and 9 antinodes.

a) Determine the wavelength (λ) of the sound wave produced by the oscillating string.

b) Calculate the frequency (f) of the sound wave.

c) Based on the speed of sound in air at room temperature, find the speed (v) of the sound wave.

---

Answer:

a) To determine the wavelength (λ) of the sound wave produced by the oscillating string, we need to analyze the number of nodes and antinodes.

A node is a point on the string that does not move during the oscillation, while an antinode is a point on the string that undergoes maximum displacement.

In a standing wave pattern, the number of nodes and antinodes can be related to the wavelength using the following equation:

λ = (2L) / n

where λ is the wavelength, L is the length of the string, and n is the number of nodes or antinodes.

In this case, we are given that there are 8 nodes and 9 antinodes. Since there are more antinodes than nodes, we will use the equation based on the number of antinodes:

λ = (2L) / n = (2 * 0.50 m) / 9 ≈ 0.111 m

Therefore, the wavelength of the sound wave produced by the oscillating string is approximately 0.111 m.

b) To calculate the frequency (f) of the sound wave, we can use the equation:

v = fλ

where v is the velocity of the sound wave, f is the frequency, and λ is the wavelength.

Since the velocity of the sound wave is not given, we need to consider that the wave is produced on the string, and thus the velocity is equivalent to the wave speed on the string. The wave speed on a string is given by:

v = √(T / μ)

where T is the tension in the string and μ is the mass per unit length of the string.

From the given information, the tension in the string is 80 N. The string length is 0.50 m, but we need to find the mass per unit length (μ) to calculate the wave speed.

To determine μ, we need to consider the linear mass density of the string (μ₀):

μ₀ = m / L

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

Since the tension is given and the string is fixed at both ends, the total mass of the string can be determined by:

μ₀ = (√(T / μ₀))² μ₀ = T / v²

Rearranging the equation, we can solve for μ₀:

μ₀ = T / v² v² = T / μ₀ v = √(T / μ₀)

Now, let's substitute values to find v:

v = √(80 N / μ₀)

We need to calculate μ₀, which can be done by considering the density (ρ) of the string material and the cross-sectional area (A):

μ₀ = ρ * A

Assuming the density and cross-sectional area are uniform, we can write μ₀ in terms of the mass per unit length (μ₁):

μ₀ = μ₁ / A

Substituting this equation into the expression for v:

v = √(80 N / (μ₁ / A))

The cross-sectional area cancels out, giving us:

v = √(80 N * (1 / μ₁)) v = √(80 N / μ₁)

Now, we need to find μ₁, the mass per unit length of the string, which can be determined as:

μ₁ = m₁ / L

To calculate μ₁, we need to find m₁. The total mass of the string (m) can be calculated using:

m = ρ * V

where ρ is the density of the string material and V is the volume of the string. Assuming the string is cylindrical in shape, we can write the volume as:

V = A * L

Substituting this equation for V into the expression for m:

m = ρ * (A * L) m = ρ * μ₀ * L

Now we have the total mass of the string, which we can use to calculate the mass per unit length (μ₁):

μ₁ = m / L μ₁ = (ρ * μ₀ * L) / L μ₁ = ρ * μ₀

Substituting this into the expression for v:

v = √(80 N / μ₁) v = √(80 N / (ρ * μ₀))

Now, let's substitute the given information to calculate the wave speed on the string:

Given: L = 0.50 m, T = 80 N

v = √(80 N / (ρ * μ₀))

Since the density of the string material (ρ) is not given, we cannot calculate the exact value of the wave speed. However, based on the speed of sound in air at room temperature, we can assume that the wave speed on the string is approximately the speed of sound in air.

c) The speed of sound in air at room temperature is approximately 343 m/s. Therefore, we can assume that the speed (v) of the sound wave on the oscillating string is approximately 343 m/s.