TS EAMCET 5-Aug-2021 Shift 1 Question Paper

Given, frequency of second wave, f2=200Hz
Speed of sound in air, v=340ms−1
Distance between two point, ∆x=85cm
As we know that,
∆φ=k∆x . . . (i)
where, ∆φ is phase difference, k is propagation constant and ∆x is path difference.
∴‌‌ωt=k∆x
where, ω is angular frequency and t is time taken.
k=‌

ω

v

‌‌(v=‌

∆x

t

)
where, v is wave velocity.
Substituting the value in Eq. (i), we get
∆φ‌‌=‌

ω

v

∆x=‌

2πf2

v

∆x
‌‌=‌

2π×200

340

×‌

85

100

‌‌=314rad
‌‌=πrad