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JEE Main 202117 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
17 March
Shift
Evening Shift
Subject
Physics
Chapter
Waves and Sound

JEE Main 2021 · 17 March · Evening Shift · Physics · Waves and Sound

A sound wave of frequency 245Hz travels with the speed of 300 ms ^...

A sound wave of frequency 245Hz245Hz travels with the speed of 300ms−1300{ms}^{- 1} along the positive x-axis. Each point of the wave moves to and fro through a total distance of 6cm6cm. What will be the mathematical expression of this travelling wave?

Options

  1. AY(x,t)=0.03[sin⁡5.1x−(0.2×103)t]Y ( x, t ) = 0 . 03\left[\sin 5 . 1x - \left(0 . 2 \times {10}^{3}\right)t\right]
  2. BY(x,t)=0.06[sin⁡5.1x−(1.5×103)t]Y ( x, t ) = 0 . 06\left[\sin 5 . 1x - \left(1 . 5 \times {10}^{3}\right)t\right]
  3. CY(x,t)=0.06[sin⁡0.8x−(0.5×103)t]Y ( x, t ) = 0 . 06\left[\sin 0 . 8x - \left(0 . 5 \times {10}^{3}\right)t\right]
  4. DY(x,t)=0.03[sin⁡5.1x−(1.5×103)t]Y ( x, t ) = 0 . 03\left[\sin 5 . 1x - \left(1 . 5 \times {10}^{3}\right)t\right]

Correct answer

D. Y(x,t)=0.03[sin⁡5.1x−(1.5×103)t]Y ( x, t ) = 0 . 03\left[\sin 5 . 1x - \left(1 . 5 \times {10}^{3}\right)t\right]

Step-by-step solution

ω=2\pif\omega = 2\pif

=1.5×103= 1 . 5 \times {10}^{3}

A=62=3cm=0.03mA = \dfrac{6}{2} = 3cm = 0 . 03m

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