MHT CET · Physics · Waves and Sound
The distance between two consecutive points with phase difference of \(60^{\circ}\) in wave of frequency 500 Hz is 0.6 m . The velocity with which wave is travelling is
- A \(1.8 \mathrm{~km} / \mathrm{s}\)
- B \(9 \mathrm{~km} / \mathrm{s}\)
- C \(3.6 \mathrm{~km} / \mathrm{s}\)
- D \(2.7 \mathrm{~km} / \mathrm{s}\)
Answer & Solution
Correct Answer
(A) \(1.8 \mathrm{~km} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
\(\phi=60^{\circ}=\frac{\pi}{3}\)
Path difference is given by,
\(\begin{array}{ll}
\quad \mathrm{x} & =\frac{\lambda}{2 \pi} \times \phi \\
\therefore \quad 0.6 & =\frac{\lambda}{2 \pi} \times \frac{\pi}{3} \\
\therefore \quad \lambda & =3.6 \mathrm{~m} \\
\quad v & =\mathrm{n} \lambda=500 \times 3.6=1800 \mathrm{~m} / \mathrm{s}=1.8 \mathrm{~km} / \mathrm{s}
\end{array}\)
Path difference is given by,
\(\begin{array}{ll}
\quad \mathrm{x} & =\frac{\lambda}{2 \pi} \times \phi \\
\therefore \quad 0.6 & =\frac{\lambda}{2 \pi} \times \frac{\pi}{3} \\
\therefore \quad \lambda & =3.6 \mathrm{~m} \\
\quad v & =\mathrm{n} \lambda=500 \times 3.6=1800 \mathrm{~m} / \mathrm{s}=1.8 \mathrm{~km} / \mathrm{s}
\end{array}\)
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