Question

In a sine wave, minimum distance between \(2\) particles always having same speed is

• A. \(\frac{\lambda }{2}\)
• B. \(\frac{\lambda }{4}\)
• C. \(\frac{\lambda }{3}\)
• D. \(\lambda\)

Hint:

Two particles half a wavelength apart have opposite displacements and always equal speeds.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For \(y=A\sin(kx-\omega t)\), particle speed is \(|v|=A\omega\left|\cos(kx-\omega t)\right|\).

Step 2: Condition
Two particles must have equal \(|\cos(kx-\omega t)|\) at all times, so their phases must differ by \(n\pi\).

Step 3: Smallest separation
\(k\,\Delta x=\pi\), so \(\Delta x=\dfrac{\pi}{k}=\dfrac{\lambda}{2}\).

Step 4: Compare the options
This is option (A).

Final Answer:
The minimum distance is half a wavelength. \[ \boxed{\text{(A) }\dfrac{\lambda}{2}} \]