Q11 / 117 · NTA Abhyas (JEE Main) 2020 · Physics · Waves and Sound

The phase difference between two points separated by 0.8 m in a wave of frequency 120 Hz is 0.5π. The wave velocity is

  • A
    144 m s-1
  • Correct answer
    B
    384 m s-1
  • C
    256 m s-1
  • D
    720 m s-1

Solution

Here, f=120 Hz, x=0.8 m and ϕ=0.5π.

From, ϕ=2πλx and λ=2πxϕ=2π×0.80.5π=3.2 m

v=nλ=120×3.2=384 m s-1

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

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