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NEET PHYSICSMedium

Two waves are represented by the equations y1=asin⁡(ωt+kx+0.57) my_1=a\sin(\omega t+kx+0.57)\text{ m} and y2=acos⁡(ωt+kx) my_2=a\cos(\omega t+kx)\text{ m}, where xx is in metre and tt in second. The phase difference between them is:

A

1.25 rad1.25 \text{ rad}

B

1.57 rad1.57 \text{ rad}

C

0.57 rad0.57 \text{ rad}

D

1.0 rad1.0 \text{ rad}

Step-by-Step Solution

  1. Identify the Given Wave Equations: The given equations are y1=asin⁡(ωt+kx+0.57)y_1 = a \sin(\omega t + kx + 0.57) and y2=acos⁡(ωt+kx)y_2 = a \cos(\omega t + kx).
  2. Convert to the Same Trigonometric Function: To easily compare phases, both equations should be in the same trigonometric form. We can convert y2y_2 from cosine to sine using the identity cos⁡(θ)=sin⁡(θ+π2)\cos(\theta) = \sin(\theta + \frac{\pi}{2}). y2=asin⁡(ωt+kx+π2)y_2 = a \sin\left(\omega t + kx + \frac{\pi}{2}\right)
  3. Substitute the Value of π\pi: Since π≈3.14\pi \approx 3.14, π2≈1.57 rad\frac{\pi}{2} \approx 1.57 \text{ rad}. y2=asin⁡(ωt+kx+1.57)y_2 = a \sin(\omega t + kx + 1.57)
  4. Calculate the Phase Difference: The phase of the first wave is ϕ1=ωt+kx+0.57\phi_1 = \omega t + kx + 0.57 and the phase of the second wave is ϕ2=ωt+kx+1.57\phi_2 = \omega t + kx + 1.57. The phase difference Δϕ=ϕ2−ϕ1=(ωt+kx+1.57)−(ωt+kx+0.57)=1.0 rad\Delta \phi = \phi_2 - \phi_1 = (\omega t + kx + 1.57) - (\omega t + kx + 0.57) = 1.0 \text{ rad}.
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