Question

Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted outside on a pole. In winter, when the temperature is 273 K, the diffraction angle θ has a value of 19.5°. What is the diffraction angle for the same sound on a summer day when the temperature is 313 K?

Answer 1

The diffraction angle of sound from a loudspeaker changes with temperature because the speed of sound varies with temperature. The higher the temperature, the faster the sound travels, resulting in a longer wavelength and, consequently, a larger diffraction angle.

Step-by-step explanation:

The question involves the concept of wave diffraction, specifically as it pertains to sound waves exiting a diffraction horn loudspeaker. In this scenario, the diffraction angle θ of sound is observed to change with temperature, due to the dependence of the speed of sound on air temperature. To find the new diffraction angle when the temperature rises from 273 K to 313 K, we need to understand that the speed of sound in air increases with temperature. The formula v = 331.4 + 0.6T (with T in ℃) can be used to calculate the speed of sound at different temperatures, but here we'll use Kelvin and recall that v = 331.4 + 0.6(T - 273) to scale it correctly.

At 273 K, the speed of sound is approximately 331.4 m/s, and at 313 K, it is higher. Assuming the frequency of the sound remains constant, and using the relationship v = fλ (where v is the speed of sound, f is the frequency, and λ is the wavelength), the wavelength will increase with the speed of sound. According to the diffraction formula for a single slit, sin(θ) = mλ/d (where m is the order of the minimum, λ is the wavelength, and d is the width of the slit), if wavelength λ increases while d stays the same, the angle θ must also increase to satisfy the equation for the first minimum (m=1). Thus, the diffraction angle on a summer day when the temperature is higher would be larger than 19.5°.

Answer 2

`\theta = 20.98 degree`

Step-by-step explanation:

As we know that the speed of the sound is given as

`v = 332 + 0.6 t`

now at t = 273 k = 0 degree

`v = 332 m/s`

so we have

`a sin\theta = N\lambda`

`a sin\theta = N((v_1)/(f))`

now when temperature is changed to 313 K we have

`t = 313 - 273 = 40 degree`

now we have

`v = 332 + (0.6)(40)`

`v_2 = 356 m/s`

`a sin\theta' = N((v_2)/(f))`

now from two equations we have

`(sin19.5)/(sin\theta) = (332)/(356)`

so we have

`sin\theta = 0.358`

`\theta = 20.98 degree`