Stationary waves of frequency 300 Hz are formed in a string, in which the velocity of wave is 200 m/s. The distance between a node and the consecutive antinode is
A
16 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
32 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A16 m Given, f=300 Hz
Velocity of transverse wave is v=fλ ⇒v=300×λ ⇒200=300×λ ∴λ=23m
Distance between a node and consecutive antinode is Δx=λ4 ∴Δx=212=16m.