Answer to Question #195520 in Mechanics | Relativity for Mohammad Balhas

Question

The distance between the first and the third nodes of a standing wave is 0.2 m, its maximum displacement is 0.02 m and its frequency is 40 Hz. The wave-functions of the two travelling waves which interfere to give the standing wave are then:

y1 = 0.02 sin⁡(20πx-320πt) ; y2 = 0.02 sin⁡(20πx+320πt),

y1 = 0.01 sin⁡(5πx-40πt) ; y2 = 0.01 sin⁡(5πx+40πt),

y1 = 0.01 sin⁡(10πx-80πt) ; y2 = 0.01 sin⁡(10πx+80πt),

y1 = 0.005 sin⁡(5πx-40πt) ; y2 = 0.005 sin⁡(5πx+40πt),

y1 = 0.04 sin⁡(20πx-320πt) ; y2 = 0.04 sin⁡(20πx+320πt),

Answer 1

from the given information we know that:

maximum displacement =amplitude of standing wave=2A=0.02

$\omega$ (angular frequency)=$2\pi f$ =$80\pi$

k=$10\pi$

we get standing wave as:

$0.04sin(10\pi x)sin(80\pi t)$

using formula

2sinC SinD= Sin(C-D)+sin(C+D)

we get equations

(option C)

y1 = 0.01 sin⁡(10πx-80πt) ;

y2 = 0.01 sin⁡(10πx+80πt),