Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

A point isotropic sonic source of sound power 1 milliwatts emits sound of frequency 170Hz in all directions.
A. Find the distance of a point from the source where the loudness level is60dB. [Take vS=340ms1v]
B. It is known that at some moment t=0, the displacement of air particles at a certain point 4m away from the source is A. The amplitude of oscillation at this point is known to be 2A. Find the displacement of air particles in terms of A, at a point 55m away from the source at the moment t=0. Consider displacement of particles to be positive if it is away from the source.

Answer
VerifiedVerified
372.9k+ views
like imagedislike image
Hint: Power (P) and intensity (I) of a sound wave are related as I=P4πr2. The loudness of sound is β=10log10(II0).
To calculate the displacement, use the general displacement formula S=Smaxcos(kxωt+ϕ).
Where, Smax is the maximum displacement of air particles or amplitude.
k=2πλ
ω=2πf
ϕ is the initial phase
λ is the wavelength of the sound wave and it is defined as λ=vf

Complete step by step solution:
(A). It is given that,
Power of the sound source is P=1milliwatts=103W.
Frequency of the sound, f=170Hz=170sec1.
Velocity of the sound, v=340ms1
Loudness level β=60dB
We know that the threshold Intensity of sound for human hearing is I0=1012W/m2.
Also, β=10log10(II0).
Substitute the values of β and I0 in the above formula. We got
60=10log10(I1012)
Further simplifying,
6=log10(I)log10(1012)
Further calculating the logarithmic values, we got
log10(I)=612
log10(I)=6
I=106
We know that the intensity of sound is related to its power and it is given as,
I=P4πr2.
Where, r is the distance of a point from the source.
Substitute all the values of I and P in the above formula of intensity.
106=1034πr2
Further calculating, we got
r2=1034π
r=8.92m
Hence, the distance of the point from the source is 8.92m.
(B) Let’s consider the displacement of air particles at a distance x from the source at any time t is given as,
 S=Smaxcos(kxωt+ϕ) …… (1)
Where,
Smax is the maximum displacement of air particles or amplitude.
k=2πλ
ω=2πf
ϕ is the initial phase
λ is the wavelength of the sound wave and it is given by
λ=vf
Substitute the required values in the above formula
λ=340170m
 λ=2m
So, k=2π2=π
It is given that for x=4m and t=0, the displacement is S=A and amplitude Smax=2A
From equation (1) we got,
A=2Acos(4π+ϕ)
Further calculating for ϕ,
ϕ=π3
For a distance at x=55m and time at t=0, the displacement of the particle is given as,
S=2Acos(55π+π3)
Further calculating
S=2A(12)
Or S=A
Since it is asked to consider the displacement as positive.
S=A
Hence, the displacement of air particles is A at a distance 55m from the source.

Note:
The lowest possible sound that the typical human ear can detect has an intensity of 1012W/m2. Hence, the loudness at this intensity is 0dB. The particles in air vibrate parallel to the direction of propagation of sound waves. So, the sound wave is known as longitudinal wave.
Explore Offline
Centres