Question:

A sound wave with a frequency of 1000 Hz is propagating through air. The speed of sound in air is approximately 343 m/s.

a) Calculate the wavelength of this sound wave.

b) The sound wave encounters a medium with a higher density and speed of sound, causing its frequency to change to 1500 Hz. Calculate the new wavelength of the sound wave in the new medium.

c) If the initial sound wave had an intensity level of 60 dB, determine its new intensity level after encountering the new medium. (Assume there is no reflection or absorption of the sound wave.)

Answer:

a) The speed of sound, v, is defined as the product of the wavelength, λ, and the frequency, f:

v = λf

Rearranging the equation, we can solve for the wavelength:

λ = v / f

Plugging in the given values:

λ = 343 m/s / 1000 Hz

λ ≈ 0.343 m

So, the wavelength of the sound wave is approximately 0.343 m.

b) When the sound wave encounters the new medium, its frequency changes to 1500 Hz. The speed of sound in the new medium is unknown, so let's denote it as v'. We can use the same equation as before to solve for the new wavelength:

λ' = v' / f'

We can equate the two wavelengths to find the relationship between v and v':

λ = λ' 343 m/s / 1000 Hz = λ' / 1500 Hz

Simplifying the equation:

λ' = (343 m/s * 1500 Hz) / 1000 Hz

λ' = 514.5 m / s

Therefore, the new wavelength of the sound wave in the new medium is approximately 514.5 m.

c) The intensity level, I, is given in decibels (dB) and is calculated using the formula:

I = 10 log10(I/I₀)

where I₀ is the reference intensity (10^(-12) W/m²).

To determine the new intensity level after the sound wave encounters the new medium, we need to calculate the corresponding intensity, I', using the known frequency and the formula for intensity:

I' = 10 log10(I₀ * (f'/f)^2)

Using the given values and the new frequency:

I' = 10 log10(10^(-12) * (1500 Hz / 1000 Hz)^2)

Simplifying the equation:

I' = 10 log10(10^(-12) * 1.5^2)

I' = 10 log10(2.25 * 10^(-12))

I' = 10 log10(2.25) + 10 log10(10^(-12))

I' ≈ 10 log10(2.25) + (-12)

Since log10(2.25) ≈ 0.35:

I' ≈ 10 * 0.35 - 12

I' ≈ 3.5 - 12

I' ≈ -8.5 dB

Thus, the new intensity level of the sound wave after encountering the new medium is approximately -8.5 dB.