Q139 / 160 · TS EAMCET 2024 (10 May Shift 2) · Physics

The path difference between two particles of a sound wave is 50 cm and the phase difference between them is $1.8 \pi$. If the speed of sound in air is $340 \mathrm{~ms}^{-1}$, the frequency of the sound wave is

  • A
    672 Hz
  • B
    306 Hz
  • Correct answer
    C
    612 Hz
  • D
    340 Hz

Solution

For sound wave, $\Delta x=50 \mathrm{~cm}, \Delta \phi=1.8 \pi$ $\therefore$ Path difference, $\Delta x=\frac{\lambda}{2 \pi} \cdot \Delta \phi$ $\Rightarrow \quad 50=\frac{\lambda}{2 \pi} \times 1.8 \pi \therefore \lambda=\frac{500}{9} \mathrm{~cm}$ $\therefore \quad$ Frequency, $f=\frac{v}{\lambda}=\frac{340 \times 9}{500 \times 10^{-2}}=612 \mathrm{~Hz}$

Hence, the correct answer is option: 'C'.

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