Question 470: Waves

Question

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

Accepted Answer

To find the frequency of the wave, we need to start by analyzing the given displacement equation of the wave:
$$y = C \sin \left( \frac{2 \pi}{\lambda} (at - x) \right)$$
Where:
$$y$$ is the displacement of the wave
$$C$$ is the amplitude
$$\frac{2 \pi}{\lambda}$$ is the wave number (denoting how many wavelengths fit into a unit length)
$$\lambda$$ is the wavelength
$$a$$ is some constant (likely representing the speed of the wave)
$$t$$ is time
$$x$$ is the distance
By comparing with the standard form of a travelling wave, we have:
$$y = C \sin \left( k (at - x) \right)$$
Where $$k$$ is the wave number:
$$k = \frac{2 \pi}{\lambda}$$
From the standard wave equation, the argument of the sine function is usually written as:
$$k (at - x)$$
This implies that $$ka$$ in the wave equation represents the angular frequency $$\omega$$ of the wave:
$$\omega = k a$$
Substituting $$k = \frac{2 \pi}{\lambda}$$ into $$\omega$$:
$$\omega = \left( \frac{2 \pi}{\lambda} \right) a = \frac{2 \pi a}{\lambda}$$
Angular frequency $$\omega$$ is related to the frequency $$f$$ by:
$$\omega = 2 \pi f$$
So:
$$2 \pi f = \frac{2 \pi a}{\lambda}$$
Solving for the frequency $$f$$:
$$f = \frac{a}{\lambda}$$
Therefore, the correct answer is:
Option D: $$\frac{a}{\lambda}$$

Answer

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

Answer

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

Answer

$$\frac{\lambda}{a}$$

Answer

$$\frac{a}{\lambda}$$

Waves

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

Choose the correct answer: