. For the stationary wave y = 4sin(x/15) sin100t the distance between a node and the next antinode is
For the stationary wave y = 4sin(.jpg)
x/15) sin100.jpg)
t the distance between a node and the next antinode is
x/15) sin100t the distance between a node and the next antinode is - (A):7.5
- (B):15
- (C):22.5
- (D):30
Best Answer
Correct Answer:
(A): 7.5Related questions
