MHT CET · Physics · Waves and Sound
A progressive wave is given by, \(Y=12 \sin (5 t-4 x)\). On this wave, how far away are the two points having a phase difference of \(90^{\circ}\) ?
- A \(\frac{\pi}{4}\)
- B \(\frac{\pi}{8}\)
- C \(\frac{\pi}{16}\)
- D \(\frac{\pi}{32}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{8}\)
Step-by-step Solution
Detailed explanation
Given, \(y=12 \sin (5 t-4 x)\) \(\therefore \quad y=12 \sin 2 \pi\left(\frac{5 t}{2 \pi}-\frac{4 x}{2 \pi}\right)\)
Comparing above eq. with,
\(\mathrm{y}=\mathrm{A} \sin 2 \pi\left(\frac{\mathrm{x}}{\lambda}-\frac{\mathrm{t}}{\mathrm{T}}\right)\)
We get, \(\lambda=\frac{2 \pi}{4}\)
Relation between phase difference and path difference is
\(\begin{aligned} \Delta \phi & =\frac{2 \pi}{\lambda} \Delta \mathrm{x} \\ \therefore \quad \frac{\pi}{2} & =\frac{2 \pi}{\left(\frac{2 \pi}{4}\right)} \Delta \mathrm{x} \\ \therefore \quad \Delta \mathrm{x} & =\frac{\pi}{8}\end{aligned}\)
Comparing above eq. with,
\(\mathrm{y}=\mathrm{A} \sin 2 \pi\left(\frac{\mathrm{x}}{\lambda}-\frac{\mathrm{t}}{\mathrm{T}}\right)\)
We get, \(\lambda=\frac{2 \pi}{4}\)
Relation between phase difference and path difference is
\(\begin{aligned} \Delta \phi & =\frac{2 \pi}{\lambda} \Delta \mathrm{x} \\ \therefore \quad \frac{\pi}{2} & =\frac{2 \pi}{\left(\frac{2 \pi}{4}\right)} \Delta \mathrm{x} \\ \therefore \quad \Delta \mathrm{x} & =\frac{\pi}{8}\end{aligned}\)
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