Step 1: Identify the Wave ParametersThe given wave equation is:
\[ y = \sin(20\pi x + 10\pi t) \]
The general form of a traveling wave is:
\[ y = \sin(kx + \omega t + \phi) \]
Comparing the given equation with the general form:
- Wave number (\( k \)) = \( 20\pi \, \text{rad/m} \)
- Angular frequency (\( \omega \)) = \( 10\pi \, \text{rad/s} \)
Step 2: Determine the Wavelength (\( \lambda \))The wavelength is related to the wave number by:
\[ k = \frac{2\pi}{\lambda} \]
\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{20\pi} = 0.1 \, \text{m} = 10 \, \text{cm} \]
\subsection{Step 3: Find the Oscillating Speed (\( v \))}
The oscillating speed is the time derivative of the displacement:
\[ v = \dv{y}{t} = \dv{t} \sin(20\pi x + 10\pi t) \]
\[ v = 10\pi \cos(20\pi x + 10\pi t) \]
Step 4: Condition for Same Oscillating SpeedFor two points to have the same oscillating speed at any instant, their phase angles must satisfy:
\[ \cos(\theta_1) = \cos(\theta_2) \]
This implies:
\[ \theta_2 = \theta_1 + 2n\pi \quad \text{or} \quad \theta_2 = -\theta_1 + 2n\pi \]
Given \( \theta = 20\pi x + 10\pi t \), the phase difference \( \Delta \theta \) must satisfy:
\[ \Delta \theta = 20\pi \Delta x = 2n\pi \quad \text{or} \quad \Delta \theta = 20\pi \Delta x = -2\theta_1 + 2n\pi \]
The smallest non-zero distance occurs when:
\[ 20\pi \Delta x = \pi \]
\[ \Delta x = \frac{\pi}{20\pi} = \frac{1}{20} \, \text{m} = 5 \, \text{cm} \]
\subsection{Final Answer}
The minimum distance between two points with the same oscillating speed is \(5.0 \, \text{cm}\).