JEE Main 17 March 2021 Shift 2 Solved Paper

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=‌

ω

v

Substituting the value of v and ω, we get
k=‌

1.54×103

300

⇒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)