wiz-icon
MyQuestionIcon
MyQuestionIcon
17
You visited us 17 times! Enjoying our articles? Unlock Full Access!
Question

A sound wave of frequency 100 Hz is travelling in air. The speed of sound in air is 350 m s1.

What is the phase difference at a given instant between two points separated by a distance of 10.0 cm along the direction of propagation?


A

π2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

2π35

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

π

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

25π2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B

2π35


General expression for pressure is time for sound is

P=P0 sin ω(txv)

P=P0sin (ωtωxv)

If some phase Φ is present initially then that gets added up

P=P0 sin (ωtωxv+Φ)

Lets say Φ was 0 when particle was at X0 at time to

So equation become P=P0 sin (ωt0ωx0v)

Now the same equation for a particle at a point 10 cm from x0 at same time to

P=P0 sin (ωt0ω(x0+0.1)v)

P=P0 sin (ωt0ωx0vω0.1v)

So here phase is Φ=ω0.1v

Given f=100 Hz

ω=2πf=200π

v=350 m/s

Φ=200π×1×101350=2π35

OR

2π

Phase change is given by ΔΦ=kx=2πxλ

λ=vf

350100=3.5 m

Therefore, ΔΦ=2π(0.1)3.5=2π35


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Pressure Wave
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon