Question

Transverse waves on a string have wave speed $8.00$ m/s, amplitude $0.0700m$ and wavelength $0.32m$. The waves travel in the negative x-direction and $t=0$ the $x=0$ end of the string has its maximum upward displacement. Write a wave function describing the wave.

• A. $y(x,t)=\left(0.07m\right)sin2\pi (\frac{x}{0.32m}+\frac{t}{0.04s})$
• B. $y(x,t)=\left(77m\right)cos2\pi (\frac{x}{0.32m}+\frac{t}{0.04s})$
• C. $y(x,t)=\left(0.7m\right)sin4\pi (\frac{x}{0.32m}+\frac{t}{0.04s})$
• D. $y(x,t)=\left(0.97m\right)sin2\pi (\frac{x}{0.32m}+\frac{t}{0.04s})$

Answer 1

The correct option is A $y(x,t)=\left(0.07m\right)sin2\pi (\frac{x}{0.32m}+\frac{t}{0.04s})$

A left travelling transverse wave given by $y=Asin(kx+\omega t)$

where wave number,$k=\frac{2\pi }{\lambda }=\frac{2\pi }{0.32}rad/m$

Speed of wave=$\lambda \nu =8m/s$

$⟹\nu =\frac{8}{0.32}Hz=25Hz$

$⟹\omega =2\pi \nu =\frac{2\pi }{0.04s}$

Thus $y=\left(0.07m\right)sin2\pi (\frac{x}{0.32m}+\frac{t}{0.04s})$