Question

Equation of two simple harmonic waves are given by \(Y_1 = 2 \sin 8\pi \left( \frac{t}{0.2} - \frac{x}{2} \right) m\) and \(Y_2 = 4 \sin 8\pi \left( \frac{t}{0.16} - \frac{x}{1.6} \right) m\) then both waves have

• A. same period
• B. same frequency
• C. same wavelength
• D. same velocity

Hint:

Wave velocity \(v = \frac{\omega}{k}\). Compare \(\omega/k\) for both waves. If equal, they have same velocity.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We have two wave equations in the form \(Y = A \sin \left( \frac{2\pi t}{T} - \frac{2\pi x}{\lambda} \right)\). Compare to find period, frequency, wavelength, velocity.

Step 2: Key Formula or Approach:
Write \(Y_1 = 2 \sin \left( 8\pi \cdot \frac{t}{0.2} - 8\pi \cdot \frac{x}{2} \right) = 2 \sin \left( \frac{8\pi}{0.2}t - \frac{8\pi}{2}x \right) = 2 \sin \left( 40\pi t - 4\pi x \right)\).
\(Y_2 = 4 \sin \left( 8\pi \cdot \frac{t}{0.16} - 8\pi \cdot \frac{x}{1.6} \right) = 4 \sin \left( \frac{8\pi}{0.16}t - \frac{8\pi}{1.6}x \right) = 4 \sin \left( 50\pi t - 5\pi x \right)\).
General form: \(Y = A \sin(\omega t - kx)\) where \(\omega = 2\pi f\), \(k = \frac{2\pi}{\lambda}\).
For \(Y_1\): \(\omega_1 = 40\pi \Rightarrow f_1 = 20\) Hz, \(k_1 = 4\pi \Rightarrow \lambda_1 = \frac{2\pi}{4\pi} = 0.5\) m, velocity \(v_1 = f_1\lambda_1 = 20 \times 0.5 = 10\) m/s.
For \(Y_2\): \(\omega_2 = 50\pi \Rightarrow f_2 = 25\) Hz, \(k_2 = 5\pi \Rightarrow \lambda_2 = \frac{2\pi}{5\pi} = 0.4\) m, velocity \(v_2 = 25 \times 0.4 = 10\) m/s.
Thus velocities are same.

Step 4: Final Answer:
Option (D) is correct.