For the harmonic travelling wave, y=2cos2π10t-0.008x+3.5 where x and y are in cm and t is in second. What is the phase difference between the oscillatory motion at two points separated by a distance of:

1. 4m

2. 0.5m

3. λ2

4. 3λ4 (at a given instant of time)

5. What is the phase difference between the oscillation of a particle located at x=100cm, at t = T sec and t=5 sec?

Hint: Use the standard equation of motion.
Given, the wave equation;
                                 y=2cos2π10t-0.0080x+3.5
=2cos20πt-0.016πx+7π
Now, the standard equation of a travelling wave can be written as:
                                 y=acosωt-kx+ϕ
On comparing with the above equation, we get,
                                                   a=2cm
ω=20πrad/s
k=0.016π
Step 1: Find the phase difference in each option.
1.
Path difference=4m
Phase difference,
ϕ=2πλ×pathdifference2πλ=k
ϕ=0.016π×4×100
=6.4πrad
2. 
ϕ=2πλ×0.5×100Pathdifference=0.5m
=0.016π×0.5×100
=0.8πrad
3. 
ϕ=2πλ×λ2=πrad
                                           Pathdifference=λ/2
4. 
ϕ=2πλ×3λ4=3π2rad
Step 2: Find the phase difference of the particle at t = T sec and t = 5 sec.
5. 
T=2πω=2π20π=110s
Atx=100cm,
t=T
ϕ1=20πT-0.016π100+7π
=20π110-1.6π+7π=2π-1.6π+7π......(i)
Again, at x=100 cm, t = 5s
                             ϕ2=20π5-0.016π100+7π
=100π-0.016×100π+7π
=100π-1.6π+7π....(ii)
 From Eqs. (i) and (ii), we get,
ϕ=phasedifference=ϕ2-ϕ1
=(100π-1.6π+7π)-(2π-1.6π+7π)
=100π-2π=98πrad