Question

If the speed of the wave shown in the figure is 330 m/s in the given medium, then the equation of the wave propagating in the positive x - direction will be -(all quantities are in MKS units)

• A. $y=0.05\text{sin}2\pi (4000t-12.5x)$
• B. $y=0.05\text{sin}2\pi (4000t-122.5x)$
• C. $y=0.05\text{sin}2\pi (3300t-10x)$
• D. $y=0.05\text{sin}2\pi (3300x-10t)$

Answer 1

The correct option is C $y=0.05\text{sin}2\pi (3300t-10x)$

The length $0.25m$ corresponds to $2.5\lambda$
$0.25=2.5\lambda$
$\lambda =0.1m$
$k=\frac{2\pi }{0.1}$
we know
$V=\lambda f$
here $v=$ speed of wave,
$\lambda$ = wavelength
$f=$ frequency
$330=0.1f$$\Rightarrow f=3300\Rightarrow \omega =2\pi \times 3300$

$y=\text{A sin}(\omega t-kx)$
$y=0.05\text{sin}(3300\times 2\pi t-\frac{2\pi }{0.1}x)$
$y=0.05\text{sin}2\pi (3300t-10x)$