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 = 0.01 sin(10tx-80nt) ; y2 = 0.01 sin(10tx+80nt), O y1 = 0.02 sin(20tx-320nt); y2 = 0.02 sin(20rtx+320nt), O y1 = 0.01 sin(5rx-40rt); y2 = 0.01 sin(5rtx+40nt), y1 = 0.04 sin(20rtx-320nt); y2 = 0.04 sin(20tx+320rt), y1 = 0.005 sin(5rx-40nt) ; y2 = 0.005 sin(5tx+40nt),
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 = 0.01 sin(10tx-80nt) ; y2 = 0.01 sin(10tx+80nt), O y1 = 0.02 sin(20tx-320nt); y2 = 0.02 sin(20rtx+320nt), O y1 = 0.01 sin(5rx-40rt); y2 = 0.01 sin(5rtx+40nt), y1 = 0.04 sin(20rtx-320nt); y2 = 0.04 sin(20tx+320rt), y1 = 0.005 sin(5rx-40nt) ; y2 = 0.005 sin(5tx+40nt),
Related questions
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 = 0.01 sin(10πx-80πt) ; y2 = 0.01 sin(10πx+80πt),
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 = 0.04 sin(20πx-320πt) ; y2 = 0.04 sin(20πx+320πt),
y1 = 0.005 sin(5πx-40πt) ; y2 = 0.005 sin(5πx+40πt),
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps