MHT CET20226 Aug 2022Evening ShiftPhysicsWaves and SoundActual
For a stationary wave, Y=10 ( x 15 ) (48 t) cm , the distance between a node and the successive antinode is
Options
- A7.5 ~cm
- B30 ~cm
- C15 ~cm
- D60 ~cm
Correct answer
A. 7.5 ~cm
Step-by-step solution
Comparing given equation with standard equation y=2 a ( 2 x ) (2 f t) where, a is the amplitude, k= 2 the propagation constant and f the frequency. Therefore, we obtain: 2 = 15 ~cm ⁻¹ =30 ~cm Distance between nearest node and antinode = 4 = 30 4 =7.5 ~cm