Answer to Question #178278 in Mechanics | Relativity for Cornelius Abel

Question

Standing waves on the wire are described by the expression y(x,t)=Asin (kx)sinwt. Supposing that the standing wave amplitude is given to be 2.5mm whose wave number(k) is 0.750pi and the angular frequency(w) is 942rad/s. If the left end of the wire is at x=0. At what distance from the left end are the

(1) Nodes of the standing wave?

(2) Antinodes of the standing wave?

Answer 1

Solution.

$y(x,t)=Asin (kx)sinwt;$

$A=2.5\sdot10^{-3}m;$

$k=0.750\pi (rad/m);$

$\omega= 942rad/s;$

$x=0;$

$1)sin(kx)=0\implies kx=\pi n, n=0,1,2,3...$

$x=\dfrac{\pi n}{k};$ $x=0; x=\dfrac{\pi}{0.750\pi}=1.33m; x=\dfrac{2\pi}{0.750\pi}=2.66m;x=\dfrac{3\pi}{0.750\pi}=3.99m.$;

$2)sin(kx)=1 or -1 \implies kx=\dfrac{ \pi}{2}+\pi n, n=0,1,2,3...$

$x=\dfrac{\pi}{2k}+\dfrac{\pi n}{k};$ $x=\dfrac{\pi}{2\sdot0.750\pi}=0.67m; x=2m; x=3.34m.$

Answer: $1)x=0; x=1.33m; x=2.66m;x=3.99m;$

$2)$$x=0.67m; x=2m; x=3.34m.$