Question

A sinusoidal wave moving along a string is shown twice in the figure, as crest $A$ travels in the positive direction of an $x$ axis by distance $d=6.0cm$ in $4.0m$. The tick marks along the axis are separated by $10cm$; height $H=6.00mm$. The wave equation is

• A. $y=(3mm)\sin \left[16x-2.4\times 10^{2}t\right]$
• B. $y=(3mm)\sin \left[16x+2.4\times 10^{2}t\right]$
• C. $y=(3mm)\sin \left[8x+2.4\times 10^{2}t\right]$
• D. $y=(3mm)\sin \left[8x-2.4\times 10^{2}t\right]$

Answer 1

Answer: (A)

The amplitude $A$ is half of the $6.00mm$ vertical range shown in the figure, that is, $A=3.0mm$.
The speed of the wave is $v=d/t=15m/s$,
where $d=0.060m$ and $t=0.0040s$.
The angular wave number is $k=2\pi /\lambda$ where $\lambda =0.40m$.
Thus, $k=\frac{2\pi }{\lambda }=16rad/m$.
The angular frequency is found from
$\omega =kv=(16rad/m)(15m/s)=2.4\times 10^{2}rad/s$
We choose the minus sign (between $kx$ and $\omega t$ ) in the argument of the sine function because the wave is shown traveling to the right (in the $+x$ direction). Therefore, with SI units understood, we obtain
$y=A\sin [kx-\omega t]=(3mm)\sin \left[16x-2.4\times 10^{2}t\right]$