Step 1: Understanding the Concept:
A transverse wave involves the oscillation of particles of the medium in a direction perpendicular to the direction of wave propagation.
The wave equation describes the displacement \(y\) of a particle at position \(x\) and time \(t\).
There are two distinct velocities involved in wave motion:
1. Wave Velocity (\(v_w\)): The speed at which the wave profile or energy travels through the medium. It is constant for a given medium and frequency.
2. Particle Velocity (\(v_p\)): The velocity of an individual particle of the medium as it executes simple harmonic motion about its equilibrium position. This velocity varies with time and position.
The particle velocity is found by taking the partial derivative of the displacement equation with respect to time.
Step 2: Key Formula or Approach:
General wave equation: \(y = A \sin(\omega t - kx)\).
Wave velocity: \(v_w = \frac{\omega}{k} = f \lambda\).
Particle velocity: \(v_p = \frac{\partial y}{\partial t} = \omega A \cos(\omega t - kx)\).
Maximum particle velocity: \((v_p)_{max} = \omega A\).
Step 3: Detailed Explanation:
The given wave equation is:
\[ y = y_0 \sin 2\pi(ft - \frac{x}{\lambda}) \]
Comparing this with the standard form \(y = y_0 \sin(\omega t - kx)\):
The amplitude is \(A = y_0\).
The angular frequency is \(\omega = 2\pi f\).
The wave number is \(k = \frac{2\pi}{\lambda}\).
First, let's find the wave velocity (\(v_w\)):
\[ v_w = f \lambda \]
Next, let's find the expression for particle velocity (\(v_p\)) by differentiating \(y\) with respect to \(t\):
\[ v_p = \frac{\partial y}{\partial t} = y_0 \cdot \cos[2\pi(ft - \frac{x}{\lambda})] \cdot (2\pi f) \]
The maximum value of the cosine function is 1. Therefore, the maximum particle velocity is:
\[ (v_p)_{max} = 2\pi f y_0 \]
According to the problem, the maximum particle velocity is four times the wave velocity:
\[ (v_p)_{max} = 4 \cdot v_w \]
Substituting the expressions we derived:
\[ 2\pi f y_0 = 4 \cdot (f \lambda) \]
Since frequency \(f\) is non-zero, we can divide both sides by \(f\):
\[ 2\pi y_0 = 4\lambda \]
Solving for \(\lambda\):
\[ \lambda = \frac{2\pi y_0}{4} \]
\[ \lambda = \frac{\pi y_0}{2} \]
This matches option (B).
Step 4: Final Answer:
By equating the maximum particle velocity (\(\omega A\)) to four times the wave velocity (\(f\lambda\)), the relationship for wavelength is determined as \(\lambda = \frac{\pi y_0}{2}\).