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-Dominant- [34]

2 years ago

6

A pipe that is open at both ends and a tuning fork of frequency 240 Hz are being used as a fancy new thermometer, albeit one tha

t utilizes sound interference. At room temperature (20.0°C), the pipe's fundamental frequency and the tuning fork's frequency are identical and thus no beat frequency is heard between the two. When the temperature rises to 43.0°C, what beat frequency (in Hz) should be heard?

1 answer:

garri49 [273]2 years ago

8 0

Answer:

9.51 Hz

Explanation:

f₀ = frequency of tuning fork and fundamental frequency of the pipe initially = 240 Hz

T₀ = Room temperature = 20 °C

v₀ = speed of sound at room temperature

speed of sound at room temperature is given as

v₀ = 331 + (0.6) T₀ = 331 + (0.6) (20) = 343 m/s

L₀ = Length of the pipe

For open pipe, we have

v₀ = 2 L₀ f₀

343 = 2 (240) L₀

L₀ = 0.715 m

T = new temperature = 43.0°C

v = speed of sound at new temperature

speed of sound at new temperature is given as

v = 331 + (0.6) T = 331 + (0.6) (43) = 356.8 m/s

L = length of pipe = L₀ = 0.715 m

f = new fundamental frequency

For open pipe, we have

v = 2 L f

356.8 = 2 (0.715) f

f = 249.51 Hz

Δf = Beat frequency

Beat frequency is given as

Δf = f - f₀

Δf = 249.51 - 240

Δf = 9.51 Hz

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