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NEET · Physics · STD 11 - 14. waves and sound

For a travelling harmonic wave \(y(x, t)=2.0 \cos 2 \pi(10 t-0.0080 x+0.35)\), where x and y are in cm and t in s. The phase difference between oscillatory motion of two points separated by a distance of 0.5 m is :

  1. A \(0.08 \pi rad\)
  2. B \(0.8 \pi rad\)
  3. C \(8 \pi rad\)
  4. D \(0.008 \pi rad\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(0.8 \pi rad\)

Step-by-step Solution

Detailed explanation

(B) \(0.8 \pi rad\)
The given equation of the travelling harmonic wave is \(y(x, t)=2.0 \cos 2 \pi(10 t-0.0080 x+0.35)\).
Comparing this with the standard wave equation \(y(x, t)=A \cos \left(\omega t-k x+\phi_0\right)\), we get the wave number \(k=2 \pi \times 0.0080 rad / cm\).
The distance between the two points is given as \(\Delta x=0.5 m=50 cm\).
The phase difference \(\Delta \phi\) between two points separated by a distance \(\Delta x\) is given by \(\Delta \phi=k \Delta x\).
Substituting the values, we get:
\(\Delta \phi=(2 \pi \times 0.0080) \times 50\)
\(\Delta \phi=2 \pi \times 0.4=0.8 \pi rad\)
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