Question

A simple harmonic progressive wave is given by the equation \(y=a\sin 2\pi \left(nt-\frac{x}{\lambda }\right)\). If the wave velocity is equal to \(\frac{1}{6}\times\) (maximum particle velocity), then the wavelength \(\lambda\) is \(\{a=\) amplitude, \(n=\) frequency, \(t=\) time, \(y=\) displacement, \(x=\) distance\(\}\)

• A. \(\frac{\pi a}{2}\)
• B. \(\frac{\pi a}{3}\)
• C. \(\frac{3\pi a}{2}\)
• D. \(\frac{2\pi a}{3}\)

Hint:

Compare the wave velocity n times lambda with the maximum particle velocity 2 pi n a.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The wave \(y=a\sin2\pi\left(nt-\dfrac{x}{\lambda}\right)\) has wave velocity \(v=n\lambda\).

Step 2: Maximum particle velocity
\(v_p=\dfrac{\partial y}{\partial t}\), so \(v_{p,max}=2\pi na\).

Step 3: Apply the condition
\(n\lambda=\dfrac{1}{6}(2\pi na)\)
\(\lambda=\dfrac{\pi a}{3}\)

Final Answer:
The wavelength is pi a divided by 3. \[ \boxed{\text{(B) }\dfrac{\pi a}{3}} \]