Question

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

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

Answer 1

The correct option is A $y=\left(3mm\right)\sin [16x-2.4\times {10}^{2}t]$

As per the given diagram, the amplitude $A$ is half of the $6.00mm$ vertical range that is, $A=3.00mm.$
The speed of the wave is
$V=\frac{d}{t},$
where $d=0.060m.$ and $t=0.0040s.$
Therefore,
$v=\frac{0.0060}{0.0040}=15m/s$
The angular wave number is
$k=\frac{2\pi }{\lambda }$
Where $\lambda =0.40m.$
Hence, by putting the values we get,
$k=\frac{2\pi }{\lambda }=16rad/m.$
As we know,
The angular frequency is
$\omega =kv=\left(16rad/m\right)\left(15m/s\right)$
$2.4\times {10}^{2}rad/s$
We choose the minus sign (between $kx$ and $\omega t$) in the argument of the sine function because the wave is shown travelling to the right in the $+x$ direction).
Therefore, with $SI$ units understood, we obtain
$y=A\sin [kx-\omega t]$
$y=\left(3mm\right)\sin [16x-2.4\times {10}^{2}t].$
Final answer: $\left(a\right)$