Question

A transverse sinusoidal wave is generated at one end of a long horizontal string by a bar that moves the end up and down through a distance by $2.0cm.$ The motion of bar is continuous and is repeated regularly $125$ times per sec. If the distance between adjacent wave crests is observed to be $15.7cm$ and the wave is moving along +ve x-direction, and at $t=0$ , the element of the string at $x=0$ is at mean position $y=0$ and is moving downwards, the equation of the wave is best described by : $(use\pi =3.14)$

• A. $y=\left(1cm\right)\sin \left[\right(40.0rad/m)x-(785rad/s\left)t\right]$
• B. $y=\left(2cm\right)\sin \left[\right(40.0rad/m)x=(785rad/s\left)t\right]$
• C. $y=\left(1cm\right)\cos \left[\right(40.0rad/m)x-(785rad/s\left)t\right]$
• D. $y=\left(2cm\right)\cos \left[\right(40.0rad/m)x-(785rad/s\left)t\right]$

Answer 1

Answer: (A)

$y=\left(1cm\right)\sin \left[\right(40.0rad/m)x-(785rad/s\left)t\right]$

The equation of a transverse sinusoidal wave is given by:
$A\sin (kx-\omega t+\varphi )$
Here, $\omega =2\pi f=2\times 3.14\times 125=785rad/s$
$A=1cm$ since $2cm$ up and down
$k=\frac{2\pi }{x}=\frac{2\times 3.14}{0.157}=40rad/m$