Question:medium

A simple harmonic progressive wave is given by \( Y = Y_0 \sin 2\pi \left( nt - \frac{x}{\lambda} \right) \). If the wave velocity is \(\left( \frac{1}{8} \right)^{\text{th}}\) the maximum particle velocity then the wavelength is

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For a sinusoidal wave, maximum particle velocity = amplitude × angular frequency. Wave velocity = frequency × wavelength. Keep symbols distinct to avoid confusion.
Updated On: Jun 8, 2026
  • \(\frac{\pi Y_0}{2}\)
  • \(\frac{\pi Y_0}{4}\)
  • \(\frac{\pi Y_0}{8}\)
  • \(\frac{\pi Y_0}{16}\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Note down what is given.
The wave is $Y = Y_0 \sin 2\pi \left( nt - \frac{x}{\lambda} \right)$. We are told the wave speed is one eighth of the largest speed a single particle can have. We must find the wavelength.

Step 2: Write the wave speed.
For this wave the speed of the wave itself is $v = n\lambda$, where $n$ is the frequency and $\lambda$ is the wavelength.

Step 3: Write the biggest particle speed.
A particle of the medium moves up and down. Its fastest speed is $v_{p,\max} = 2\pi n Y_0$. Here $Y_0$ is the amplitude.

Step 4: Put the given condition into symbols.
Wave speed is one eighth of the top particle speed, so $n\lambda = \frac{1}{8}\left( 2\pi n Y_0 \right)$.

Step 5: Cancel and clean up.
The frequency $n$ sits on both sides, so we cancel it. We get $\lambda = \frac{1}{8}\left( 2\pi Y_0 \right) = \frac{2\pi Y_0}{8}$.

Step 6: Simplify the fraction.
$\frac{2\pi Y_0}{8} = \frac{\pi Y_0}{4}$. So the wavelength is $\frac{\pi Y_0}{4}$.
\[ \boxed{\lambda = \frac{\pi Y_0}{4}} \]
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