Speed of sound in air is `330 m//s`. Frequancy of Anoop's voice is `1000 Hz` and of Shibham's voice is `2000 Hz`. Find the wavelength corresponding to their voice.
Text Solution
Verified by Experts
The correct Answer is:
A, C
`lambda = (v)/(f)` for both
Topper's Solved these Questions
WAVE MOTION
DC PANDEY|Exercise Exercise 17.5|6 Videos
WAVE MOTION
DC PANDEY|Exercise Exercise 17.6|6 Videos
WAVE MOTION
DC PANDEY|Exercise Exercise 17.3|4 Videos
VECTORS
DC PANDEY|Exercise Medical enrances gallery|9 Videos
WORK, ENERGY & POWER
DC PANDEY|Exercise Level 2 Comprehension Based|2 Videos
Similar Questions
Explore conceptually related problems
If speed of sound in air is 340 m//s and in water is 1480 m//s .If frequency of sound is 1000 kHz, then find wavelength in water
Speed of sound in air is 330m//s .Find maximum and minimum wavelength of audible sound in air.
The frequency of a man's voice is 200 Hz and its wavelength is 2 m . If the wavelength of a child's voice in 1 m , then the frequency of the child's voice is :
Frequency of a sound wave produced by a vibrating body is 50 Hz. Find the wavelength of the sound wave produced. (V_("air") = 330 ms^(-1))
The frequency of a man's voice is 300 Hz and its wavelength is 1 meter. If the wavelength of a child's voice is 1.5 m, then the frequency of the child's voice is"
(a) A cylindrical metal tube has a length of 50 cm and is open at both ends. Find the frequencies between 1000 Hz and 2000 Hz at which the air is 340 m//s . (b) Find the greatest length of an organ pipe open at both ends that will have its fundamental frequency in the normal hearing range (20 - 20000 Hz) . Speed of sound in air = 340 m//s . (c) Two successive resonance frequencies in an open organ pipe are 1944 Hz and 2592 Hz . Find the length of the tube. The speed of sound in air is 324 m//s .
