Question:
Which of the following equations correctly represents a travelling wave having wavelength \( \lambda \) = 4.0 cm, frequency v = 100 Hz and travelling in positive x-axis direction?
Which of the following equations correctly represents a travelling wave having wavelength \( \lambda \) = 4.0 cm, frequency v = 100 Hz and travelling in positive x-axis direction?
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A wave travelling in +x has the form y = A sin(kx - wt) with k = 2pi/lambda and w = 2pi v; compare with each option.
Updated On: Oct 31, 2026
- \( y = A\sin [(0.50\,\pi \,c{m^{ - 1}})x - (100\,\pi \,{s^{ - 1}})t] \)
- \( y = A\sin \,\,2\pi [(0.25\,\,c{m^{ - 1}})x - (50\,{s^{ - 1}})t] \)
- \( y = A\sin \left[ {\left( {{{2\pi } \over 4}\,c{m^{ - 1}}} \right)x - \left( {{{2\pi } \over {100}}\,{s^{ - 1}}} \right)t} \right] \)
- \( y = A\sin \,\pi [(0.5\,\,c{m^{ - 1}})x - (200\,\,{s^{ - 1}})t] \)
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The Correct Option is D
Solution and Explanation
Step 1: \( k=\frac{2\pi}{\lambda}=\frac{2\pi}{4\ \text{cm}}=0.5\pi\ \text{cm}^{-1} \).
Step 2: \( \omega=2\pi\nu=2\pi\times100=200\pi\ \text{s}^{-1} \).
Step 3: Wave travelling in +x: \( y=A\sin(kx-\omega t)=A\sin[(0.5\pi\ \text{cm}^{-1})x-(200\pi\ \text{s}^{-1})t]=A\sin\pi[(0.5\ \text{cm}^{-1})x-(200\ \text{s}^{-1})t] \).
Step 4: This matches option (D). Option (A) has \( \omega=100\pi \) (wrong), option (B) gives \( \omega=100\pi \) (wrong), option (C) has \( \omega=\frac{2\pi}{100} \) (wrong).
Answer: (D) \( y=A\sin\pi[(0.5\,\text{cm}^{-1})x-(200\,\text{s}^{-1})t] \)
Step 2: \( \omega=2\pi\nu=2\pi\times100=200\pi\ \text{s}^{-1} \).
Step 3: Wave travelling in +x: \( y=A\sin(kx-\omega t)=A\sin[(0.5\pi\ \text{cm}^{-1})x-(200\pi\ \text{s}^{-1})t]=A\sin\pi[(0.5\ \text{cm}^{-1})x-(200\ \text{s}^{-1})t] \).
Step 4: This matches option (D). Option (A) has \( \omega=100\pi \) (wrong), option (B) gives \( \omega=100\pi \) (wrong), option (C) has \( \omega=\frac{2\pi}{100} \) (wrong).
Answer: (D) \( y=A\sin\pi[(0.5\,\text{cm}^{-1})x-(200\,\text{s}^{-1})t] \)
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