Question

A progressive wave of frequency 25 Hz is travelling through a medium. Find the phase difference between two positions of a particle at an interval of 0.01 s.
​

Answer 1

Answer:

Given :   f=25Hz,A=2.5×10

−5

m,v=300m/s,

from ,   v=fλ,

or         λ=v/f=300/25=12m,

As points A and B are 6cm apart hence path difference between A and B will be ,   Δx=6cm,

by     Δϕ=

λ

2π

​

Δx,

        Δϕ=

12

2π

​

×6=π    (phase difference between A and B),

Now, displacement  for particle A,

           y

A

​

=Asinωt,

or         y

A

​

=2.5×10

−5

sinωt,

and , displacement  for particle B,

           y

B

​

=Asin(ωt+π), (as wave is travelling in -ive x-direction)

or         y

B

​

=2.5×10

−5

sin(ωt+π),

or         y

B

​

=−2.5×10

−5

sinωt,

therefore , difference in displacements of A and B,

          y

A

​

−y

B

​

=[2.5×10

−5

−(−2.5×10

−5

)]sinωt,

this difference will be maximum when , sinωt=1(maximum),

therefore , Δ=[2.5×10

−5

−(−2.5×10

−5

)]=5×1

Explanation:

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Answer 2

Given:

f= 25 Hz

t = 0.01 s

$\phi=\frac{2 \pi}{T} \cdot \Delta t$

First find T

$T=\frac{1}{f}$

$T=\frac{1}{25}$

T= 0.04 sec

$\phi=\frac{2 \pi}{0.04}(0.01)$

$\phi=\frac{\pi}{2}$

Hence the answer is $\phi=\frac{\pi}{2}$