Equation of two simple harmonic waves is given by $\mathrm{Y}_1=2 \sin 8 \pi\left(\frac{\mathrm{t}}{0…
Equation of two simple harmonic waves is given by $\mathrm{Y}_1=2 \sin 8 \pi\left(\frac{\mathrm{t}}{0.2}-\frac{\mathrm{x}}{2}\right) \mathrm{m}$ and $\mathrm{Y}_2=4 \sin 8 \pi\left(\frac{\mathrm{t}}{0.16}-\frac{\mathrm{x}}{1.6}\right) \mathrm{m}$ then both waves have
same period
same frequency
same wavelength
same velocity
Solution
The two equations can be written as
$\begin{aligned}
& Y_1=2 \sin 2 \pi\left(\frac{4 t}{0.2}-\frac{4 x}{2}\right)=2 \sin 2 \pi\left(\frac{\mathrm{t}}{0.05}-\frac{\mathrm{x}}{0.5}\right) \\
& \text { and } Y_2=4 \sin 2 \pi\left(\frac{4 \mathrm{t}}{0.16}-\frac{4 \mathrm{x}}{1.6}\right)=2 \sin 2 \pi\left(\frac{\mathrm{t}}{0.04}-\frac{\mathrm{x}}{0.4}\right)
\end{aligned}$
Comparing with standard equation
$\mathrm{Y}=\mathrm{A} \sin 2 \pi\left(\frac{\mathrm{t}}{\mathrm{T}}-\frac{\mathrm{x}}{\lambda}\right)$
We get, for the first wave,
$\mathrm{T}=0.05 \mathrm{~s} \text { and } \lambda=0.5 \mathrm{~m}$
For the second wave,
$\mathrm{T}=0.04 \mathrm{~s} \text { and } \lambda=0.4 \mathrm{~m}$
Hence their periods (hence frequencies) are not same.
Their wavelength is also not same.
For first wave, velocity $=\frac{\lambda}{\mathrm{T}}=\frac{0.5}{0.05}=10 \mathrm{~m} / \mathrm{s}$
For second wave, velocity $\frac{0.4}{0.04}=10 \mathrm{~m} / \mathrm{s}$
Hence velocity in same