Home
Class 12
PHYSICS
A small speaker has a capacity of power ...

A small speaker has a capacity of power 3 W. A microphone is placed at distance 2 m from the speaker. Fine the displacement amplitude of particles of air near to microphone. The frequenct of sound emitted bt speaker is 1.0 KHz
(Density of air = 1.2 `kg//m^(3)` and speed of sound in air = 330 m/s)

A

`2.76xx10^(-4)cm`

B

`4xx10^(-4)cm`

C

`10xx10^(-4)cm`

D

`3.8xx10^(-3)cm`

Text Solution

Verified by Experts

The correct Answer is:
A
`becauselp/(4pir^(2))=3/(4pixx2^(2))=3/(16pi)`
But `l=2pi^(2) A^(2)v^(2)p_(o)c` or `A^(2)=l/(2pi^(2)v^(2)p_(o)c)`
`therefore" "A=sqrt(((l)/(2pi^(2)v^(2)p_(o)c)))=sqrt((3)/((16pixx2pi^(2)xx10^(6)xx12xx330)))`
`0.002764xx10^(-3)m=2.76xx10^(-4)cm`
Promotional Banner

Topper's Solved these Questions

  • SOUND WAVES

    BITSAT GUIDE|Exercise BITSAT Archives|1 Videos

  • SOLVED PAPER 2018

    BITSAT GUIDE|Exercise Part-I (Physics)|40 Videos

  • UNIVERSE

    BITSAT GUIDE|Exercise All Questions|30 Videos

Similar Questions

Explore conceptually related problems

Find the amplitude of the vibration of the particles of air through which a sound wave of intensity 2.0xx10^-6Wm^-2 and frequency 1.0kHz is passing. Density of air =1.2 kgm^-3 and speed of sound in air =330 ms^-1 .

Calculate the intensity of a note of frequency 1000 Hz if the amplitude of vibration is 10^(-9) cm . Density of air = 1.3 kg m^(-3) and velocity of sound = 340 m s^(-1)

A 10 W source of sound of frequency 1000 Hz sends out wave in air. The displacment amplitude a distance of 10 m from the source is (speed of sound in air = 340 m/s and density of air = 129 kg/m^(3) )

Plane harmonic waves of frequency 500 Hz are produced in air with displacement amplitude of 10mum . Given that density of air is 1.29(kg)/(m^3) and speed of sound in air is 340(m)/(s) . Then

A note of frequency 300 Hz has an intensity of 1 microwatt per square metre. What is the amplitude of the air vibrations caused by this sound ? (Density of air = 1.293 kg m^(-3) and velocity of sound in air = 332 ms^(-1) )

A typical loud sound wave with a frequency of 1 Kh_(Z) has a pressure amplitude of about 10 Pa (a) At t = 0 , the pressure is a maximum at some point X_(1) . What is the displacement at that point at t = 0 ? (b) What is the maximum value of the displacement at any time and place/ Take the density of air to be 1.29 kg//m^(3) and speed of sound in air is 340 m//s .

A sound wave having a frequency of 100 Hz and pressure amplitude of 10 pa, then calculate the displacement amplitude (Given speed od sound in air = 340 m/s and density of air = 1.29 kg/m^(3) )

If the density of air at NTP is 1.293kg//m^(3) and gamma=1.41 , then the velocity of sound in air at NTP is :

Find the displacement ammplitude amplitude of amplitude of particles of air of density 1.2 kg//m^(3) ,if intensity, frequency and speed of sound are 8 xx 10^(-6) w//m^(2), 5000 Hz and 330 m//s respectively.