Ask Question
21 July, 19:38

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 that 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?

+2
Answers (1)
  1. 21 July, 20:35
    0
    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
Know the Answer?