The correct option is
D Its propagation velocity is
50×103 cm/sec
The standard equation of a wave is given by,
y=Asin(ωt−kx)
Let's now convert the given equation into the standard wave equation.
y=4sin2π(t0.02−x100)
or, y=4sin(2π0.02t−2π100x)
Comparing this with the standard equation we get,
A=4,ω=2π0.02,k=2π100
Amplitude is 4cm which is true, so option (A) is correct.
Wavelength, λ=2πk=2π2π100=100cm. So, option (B) is also correct.
Frequency, f=ω2π=2π0.022π=50cycles/sec. So, option (C) is also correct.
The propagation velocity, v=ωk=2π0.022π100=5000cm/s=50×102cm/s. Hence option (D) is not true for this wave.
Hence Option is D