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 2.0 cm. The motion of bar is continuous and is repeated regularly 125 times per second. If the distance between adjacent wave crests is observed to the 15.6 cm and the wave is moving along positive x-direction, and at t = 0 the element of the string at x = 0 is at mean position y = 0 and is moving downward, the equation of the wave is best described by

• A. y = (1 cm) sin [(40.3 rad/m) x – (786 rad/s)t]
• B. y = (2 cm) sin [(40.3 rad/m) x – (786 rad/s)t]
• C. y = (1 cm) cos [(40.3 rad/m) x – (786 rad/s)t]
• D. y = (2 cm) cos [(40.3 rad/m) x – (786 rad/s)t]

Answer 1

The correct option is A y = (1 cm) sin [(40.3 rad/m) x – (786 rad/s)t]

Amplitude of wave,
$A=\frac{2.0cm}{2}=1cm$
Frequency of wave, f = 125 Hz.
Wavelength of wave, $\lambda =15.6cm=0.156m$
Let equation of wave be, $y=Asin(kx-\omega t+\varphi )wherek=\frac{2\pi }{\lambda }=40.3rad/mand\omega =2\pi f=785rads$
Using initial conditions.
$y(0,0)=0=Asin\varphi and\frac{\delta y}{\delta t}(0,0)=-A\omega cos\varphi <0weget,\varphi =2n\pi$
So, the equation of wave is
y = (1 cm) sin [(40. 3 rad/m) x – (785 rad/s)t]