Q1 / 117 · NTA Abhyas (JEE Main) 2020 · Physics · Waves and Sound

The displacement y of a particle in a medium can be expressed as y=10-6sin100t-20x+π/4 m, where t is in seconds and x in meters. The speed of the wave is

  • A
    2000 m s-1
  • Correct answer
    B
    5 m s-1
  • C
    20 m s-1
  • D
    5π m s-1

Solution

Given wave equation :

y=10-6sin100t+20x+π4 m

Standard equation : y=a sin ωt+kx+ϕ

Comparing the two, we obtain

∴   ω=100 and  k=20

∴    ωk=10020

⇒2πn2π/λ=nλ=v=5

∴   v=5 m s-1

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

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