Question

If a travelling wave is given by \( y(x, t) = 0.5 \sin(70.1x - 10\pi t) \), where \( x \) and \( y \) are in metres, the time \( t \) is in seconds, then the frequency of the wave is

• A. 6 Hz
• B. 7 Hz
• C. 4 Hz
• D. 5 Hz
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Hint:

Extract \( \omega \) from the wave equation and use \( f = \frac{\omega}{2\pi} \) to find frequency in Hz.

Answer 1

The Correct Option is D

The general form of a travelling wave is: \[ y(x, t) = A \sin(kx \pm \omega t + \phi) \] where \( A \) is the amplitude, \( k \) is the wave number (in rad/m), \( \omega \) is the angular frequency (in rad/s), and \( \phi \) is the phase constant. For the given wave: \[ y(x, t) = 0.5 \sin(70.1x - 10\pi t) \] Compare with the general form: - Amplitude \( A = 0.5 \, \text{m} \). - Wave number \( k = 70.1 \, \text{rad/m} \). - Angular frequency \( \omega = 10\pi \, \text{rad/s} \). The frequency \( f \) (in Hz) is related to the angular frequency by: \[ \omega = 2\pi f \] Solve for \( f \): \[ 10\pi = 2\pi f \implies f = \frac{10\pi}{2\pi} = 5 \, \text{Hz} \] Verify other parameters for context: - Period: \( T = \frac{1}{f} = \frac{1}{5} = 0.2 \, \text{s} \). - Wavelength: \( \lambda = \frac{2\pi}{k} = \frac{2\pi}{70.1} \approx 0.0896 \, \text{m} \). - Wave speed: \( v = f \lambda = 5 \cdot \frac{2\pi}{70.1} \approx 0.448 \, \text{m/s} \). The frequency matches option (4). Check options: - (1) 6 Hz: \( \omega = 2\pi \cdot 6 = 12\pi \), incorrect. - (2) 7 Hz: \( \omega = 2\pi \cdot 7 = 14\pi \), incorrect. - (3) 4 Hz: \( \omega = 2\pi \cdot 4 = 8\pi \), incorrect. - (4) 5 Hz: Correct. Thus, the frequency is 5 Hz, confirming option (4).