Question

The displacement of a travelling wave \( y = C \sin \left( \frac{2\pi}{\lambda} (at - x) \right) \), where \( t \) is time, \( x \) is distance and \( \lambda \) is the wavelength, all in S.I. units. Then the frequency of

• A. \( \frac{2\pi \lambda}{a}\)
• B. \( \frac{2\pi a}{\lambda}\)
• C. \( \frac{\lambda}{a}\)
• D. \( \frac{a}{\lambda}\)

Answer 1

The Correct Option is D

1. Standard form of a travelling wave:

$$ y = A \sin(kx - \omega t) $$

Where: 

• $A$ = Amplitude
• $k$ = Wave number, $k = \frac{2\pi}{\lambda}$
• $\omega$ = Angular frequency, $\omega = 2\pi f$
• $f$ = Frequency of the wave
• $\lambda$ = Wavelength

2. Given wave equation:

$$ y = C \sin\left(\frac{2\pi}{\lambda}(at - x)\right) $$

3. Rearrange to compare with standard form:

$$ y = C \sin\left(-\frac{2\pi}{\lambda}x + \frac{2\pi a}{\lambda}t\right) $$ $$ y = C \sin\left(\frac{2\pi}{\lambda}x - \frac{2\pi a}{\lambda}t\right) $$

Note: $\sin(-\theta) = -\sin(\theta)$, so the negative sign can be absorbed into amplitude. For wave form we use:

$$ y = C \sin(kx - \omega t) $$

4. Compare coefficients:

$$ k = \frac{2\pi}{\lambda} \quad \text{and} \quad \omega = \frac{2\pi a}{\lambda} $$

5. Find frequency $f$:

Using $\omega = 2\pi f$, we get:

$$ 2\pi f = \frac{2\pi a}{\lambda} $$

Divide both sides by $2\pi$:

$$ f = \frac{a}{\lambda} $$