# The wavelength of a travelling wave is 4m at a frequency of 10hertz. what time later will there be another crest at x=3m? ii) if the amplitude of the wave is 12m, write the equation of the wave?

May 14, 2016

I found: $0.1 s$ and $y \left(x , t\right) = 12 \cos \left(1.57 x - 62.83 t\right)$

#### Explanation:

I would write the equation as:
$y \left(x , t\right) = A \cos \left(k x - \omega t\right)$
where:
$A =$amplitude;
$k = \frac{2 \pi}{\lambda}$ is the wavenumber ($\lambda$ is the wavelength);
$\omega = \frac{2 \pi}{T}$ is the angular frequency ($T$ is the period).
You have everything here:
From the frquency you get the period as:
$T = \frac{1}{f} = \frac{1}{10} = 0.1 s$ that incidentally is the time to have the next crest at your position (the period represents the time to complete an oscillation).
$\lambda = 4 m$
so finally your wave (propagating in the $+ x$ direction) will be:
$\textcolor{red}{y \left(x , t\right) = 12 \cos \left(1.57 x - 62.83 t\right)}$

May 14, 2016

(i) $0.1 s$
(ii) $\psi \left(x , t\right) = 12 {e}^{i \left(\frac{\pi}{2} x - 20 \pi t\right)}$

#### Explanation:

A basic wave@ may be expressed in the following general equation
$\psi \left(x , t\right) = A {e}^{i \left(k x - \setminus \omega t\right)}$ ..........(1)
where $A$ is the maximum amplitude,
wavenumber $k = \frac{2 \pi}{\lambda}$, and
angular frequency $\omega = 2 \pi f$ .
($\lambda$ is wavelength and $f$ is the frequency of the wave).

From the given frequency of $10 H z$, we get time period $T = \frac{1}{f}$
$T = \frac{1}{10} = 0.1 s$. Also given is $\lambda = 4 m$
(i) What time later will there be another crest at $x = 3 m$.
We know that for any value of $x$, the traveling wave repeats itself as per time-period of the wave. Therefore, if crest appeared at $t = 0$; the next crest will occur at
$t = 0 + T = 0 + 0.1 = 0.1 s$

(ii) Inserting values of $A , k \mathmr{and} \omega$ in (1) we obtain
$\psi \left(x , t\right) = 12 {e}^{i \left(\frac{2 \pi}{4} x - 2 \pi \times 10 t\right)}$
$\psi \left(x , t\right) = 12 {e}^{i \left(\frac{\pi}{2} x - 20 \pi t\right)}$

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@which can be shown to be the usual sine and cosine forms using Euler's formula.
Rewriting the argument as,
$k x - \setminus \omega t = \frac{2 \pi}{\lambda} k - 2 \pi f t$, after substituting $v = f \lambda$
$= \left(\setminus \frac{2 \setminus \pi}{\setminus \lambda}\right) \left(x - v t\right)$
We see that this expression describes a vibration of wavelength $\setminus \lambda = \setminus \frac{2 \setminus \pi}{k}$ traveling along the $x$-direction with a constant phase velocity given by ${v}_{p} = \setminus \frac{\setminus \omega}{k}$.