Question

A simple harmonic progressive wave is given by \(y=A \sin 2 \pi\left(n t-\frac{x}{\lambda}\right)\). If the wave velocity is equal to \(=\frac{1}{3}\) (maximum particle velocity) then the wavelength \(\lambda\) is given by

• A. \(\frac{2\pi \text{ }\mathrm{A}}{3}\)
• B. \(\frac{\pi \mathrm{A}}{3}\)
• C. \(\frac{5\pi \text{ }\mathrm{A}}{3}\)
• D. \(\frac{7\pi \text{ }\mathrm{A}}{3}\)

Hint:

Wave speed is \(n\lambda\) and the maximum particle velocity is \(\omega A=2\pi nA\); equate them using the given ratio.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Comparing with \(y=A\sin2\pi\left(nt-\dfrac{x}{\lambda}\right)\), the wave velocity is \(v=n\lambda\).

Step 2: Maximum particle velocity
\(v_{p,\max}=A\omega=2\pi nA\).

Step 3: Apply the condition
\(n\lambda=\dfrac13(2\pi nA)\).

Step 4: Solve
\(\lambda=\dfrac{2\pi A}{3}\).

Final Answer:
The wavelength is \(\dfrac{2\pi A}{3}\). \[ \boxed{\text{(A) }\dfrac{2\pi \text{ A}}{3}} \]