Question

A plane wave has equation y=25sin (120t-4x). Find the
(I) amplitude
(II) wavelength
(III) wave velocity
(iv) frequency and period of the wave

(where y and x are in metres, t is in seconds)

Answer 1

The general equation describing a plane wave looks like:

$$y(x, t) = Asin(\omega t - kx),$$

here, $A$ is the amplitude of the plane wave, $k$ is the wavenumber, $\omega$ is the angular frequency of the plane wave.

I) As we can see from the equation above, the amplitude of the wave is:

$$A = 25 m.$$

II) We can find the wavelength from the formula:

$$k = \frac{2 \pi}{\lambda},$$

here, $\lambda$ is the wavelength of the wave, $k = 4 \frac{rad}{m}$ is the wavenumber.

Then, we get:

$$\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{4 \dfrac{rad}{m}} = 1.57 m.$$

III) We can find the velocity of the wave from the wave speed formula:

$$v = f \lambda,$$

here, $f$ is the frequency of the wave.

We can find the frequency of the wave from the formula:

$$\omega = 2 \pi f,$$ $$f = \dfrac{\omega}{2 \pi}.$$

Then, substituting the freuency into the ave speed formula, we can find the velocity of the wave:

$$v = f \lambda = \dfrac{\omega \lambda}{2 \pi} = \dfrac{120 \dfrac{rad}{s} \cdot 1.57 m}{2 \pi} = 30 \dfrac{m}{s}.$$

IV) We can find the frequency of the wave from the formula:

$$f = \dfrac{\omega}{2 \pi} = \dfrac{120 \dfrac{rad}{s}}{2 \pi} = 19.1 Hz.$$

We can find the period of the ave from the formula:

$$T = \dfrac{1}{f} = \dfrac{1}{19.1 Hz} = 0.052 s.$$

Answer:

I) $A = 25 m.$

II) $\lambda = 1.57 m.$

III) $v = 30 \dfrac{m}{s}.$

IV) $f = 19.1 Hz$, $T = 0.052 s.$