Question

A transverse wave on a string is modeled with the wave function y(x, t) (0.80 m)sin[(0.85 m)x (1.70 s)t 0.20]. (Indicate the direction with the signs of your answers.) (a) Find the wave velocity (in m/s). m/s (b) Find the position (in cm) in the y-direction, the velocity (in cm/s) perpendicular to the motior of the wave, and the acceleration (in cm/s2) perpendicular to the motion of the wave of a small segment of the string centered at x 0.40 m at time t- 5.00 s position velocity acceleration cm cm/s cm/s2

Answer 1

Given wave equation for displacement is $y(x,t)=(0.80)\sin(0.85 x+1.70t+0.20)$

$k=0.85\,m^{-1}$ , $\omega=1.70\,s^{-1}$

a)

Wave velocity $v=\frac{\omega}{k}=\frac{1.70}{0.85}=2\,m/s$

b)

$y(x,t)=(0.80)\sin(0.85 x+1.70t+0.20)$

At $x=0.40\,m$ and $t=5.00\,s$

Position  $y=(0.80)\sin(0.85 (0.40)+1.70(5.00)+0.20)=0.30\,m=30\,cm$

Velocity $v(x,t)=\frac{d}{dt}y(x,t)=(0.80)(1.70)\cos(0.85 x+1.70t+0.20)$

At $x=0.40\,m$ and $t=5.00\,s$

$v=(0.80)(1.70)\cos(0.85 (0.40)+1.70(5.00)+0.20)=-1.26\,m/s=-126\,cm/s$

Acceleration $a(x,t)=\frac{d}{dt}v(x,t)=-(0.80)(1.70)^2\sin(0.85 x+1.70t+0.20)$

At $x=0.40\,m$ and $t=5.00\,s$

${a=-(0.80)(1.70)^2\sin(0.85 (0.40)+1.70(5.00)+0.20)=-0.868\,m/s=-86.8\,cm/s^2}$