Home
Class 9
PHYSICS
The distance between any two successive ...

The distance between any two successive antinodes or nodes of a stationary wave is `0.75` m . If the velocity of the wave is `300 m s^(-1)` , find the frequency of the wave .

Text Solution

Verified by Experts

The correct Answer is:
`200 Hz`

Wavelength of the waves , `lambda = 0.75 m xx 2 = 1.5` m
Velocity of the wave = `300 m s^(-1)`
frequency , `n = (v)/(lambda) = (300)/(1.5) = 200` hertz
Promotional Banner

Topper's Solved these Questions

  • WAVES MOTION AND SOUND

    PEARSON IIT JEE FOUNDATION|Exercise Test Your Concepts (Very short answer questions )|30 Videos
  • WAVES MOTION AND SOUND

    PEARSON IIT JEE FOUNDATION|Exercise Short Answer Type Questions|18 Videos
  • SIMPLE MACHINES

    PEARSON IIT JEE FOUNDATION|Exercise Level- 3|10 Videos

Similar Questions

Explore conceptually related problems

The distance between any two successive nodes or antinodes in stationary waves is

The distance between any two successive nodes or antinodes is

The distance between two successive nodes in a stationary wave is 30 cm. Calculate the frequency of wave if its speed is 250 m/s.

The distance between two consecutive nodes in a stationary wave is 25cm. If the speed of the wave is 250ms^(-1) , calculate the frequency.

The distance between any two adjacent nodes in a stationary wave is 15 cm. if the speed of the wave is 294 ms/, what is its frequency?

At nodes, the velocity of stationary wave is

The distance between two consecutive in a stationary wave is 25 cm . If the speed of the wave is 300 ms^(-1) , calculate the frequency.