Given,
The frequency of the sound wave,
f=245Hz The speed of the travelling wave,
v=300m∕s As, total distance of to and fro motion is
6cm.
Hence, the amplitude of the wave,
A=6∕2=3cm=0.03m As we know,
Angular frequency is given as
ω=2πf Substituting the value of
f, we get
ω=2π(245) ω=1.54×103rad∕s Propagation constant is given as
k= Substituting the value of
v and
ω, we get
k=⇒k=5.1m−1 General mathematical expression for a travelling wave is given as
y=Asin(kx−ωt) Substituting the values in the above equation, we get
y=0.03sin(5.1x−1.5×103t)