MHT CET · Physics · Waves and Sound
A simple harmonic progressive wave is given by \(\mathrm{Y}=\mathrm{Y}_0 \sin 2 \pi\) \(\left(\mathrm{nt}-\frac{\mathrm{x}}{\lambda}\right)\). If the wave velocity is \(\left(\frac{1}{8}\right)^{\text {th }}\) the maximum particle velocity, then the wavelength is
- A \(\frac{\pi \mathrm{Y}_0}{2}\)
- B \(\frac{\pi \mathrm{Y}_0}{4}\)
- C \(\frac{\pi \mathrm{Y}_0}{8}\)
- D \(\frac{\pi \mathrm{Y}_0}{16}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi \mathrm{Y}_0}{4}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \text { Maximum particle velocity }=\mathrm{Y}_0 \omega=2 \pi \mathrm{n} \mathrm{Y}_0 \\
& \text { Wave velocity }=\mathrm{n} \lambda \\
& \therefore \frac{2 \pi \mathrm{n} \mathrm{Y}_0}{8}=\mathrm{n} \lambda \\
& \therefore \lambda=\frac{\pi \mathrm{Y}_0}{4}
\end{aligned}
\)
\begin{aligned}
& \text { Maximum particle velocity }=\mathrm{Y}_0 \omega=2 \pi \mathrm{n} \mathrm{Y}_0 \\
& \text { Wave velocity }=\mathrm{n} \lambda \\
& \therefore \frac{2 \pi \mathrm{n} \mathrm{Y}_0}{8}=\mathrm{n} \lambda \\
& \therefore \lambda=\frac{\pi \mathrm{Y}_0}{4}
\end{aligned}
\)
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