MHT CET Medical202621 April 2026Evening ShiftPhysicsWaves and SoundActual
A string fixed at both the ends, forms a standing wave with node separation of 5 cm . If the velocity of the wave on the string is 2 m/s , then the frequency of vibration of the string is
Options
- A40 Hz
- B20 Hz
- C10 Hz
- D0.2 Hz
Correct answer
B. 20 Hz
Step-by-step solution
The distance between two consecutive nodes in a standing wave is 2 . Given that the node separation is 5 cm , we have: 2 = 5 cm = 0.05 m = 0.1 m The velocity of the wave is given as v = 2 m/s . The frequency of vibration f is related to the wave velocity and wavelength by the formula: f = v Substituting the values: f = 2 0.1 = 20 Hz Answer: 20 Hz