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Waves and Sound — JEE physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceJEE physicsWaves and Sound1 Mark
MCQ
The phase difference between two points separated by $0.8 \mathrm{~m}$ in a wave of frequency $120 \mathrm{~Hz}$ is $90^{\circ}$. Then the velocity of wave will be
  • A
    $192 \mathrm{~m} / \mathrm{s}$
  • B
    $360 \mathrm{~m} / \mathrm{s}$
  • C
    $710 \mathrm{~m} / \mathrm{s}$
  • ✓
    $384 \mathrm{~m} / \mathrm{s}$
✓

Answer

Correct option: D.
$384 \mathrm{~m} / \mathrm{s}$
(d)
 Path difference$\Delta=\frac{\lambda}{2\pi}\times\phi=\frac{\lambda}{2 \pi} \times \frac{\pi}{2}=\frac{\lambda}{4}$
$\because \Delta=0.8 \mathrm{~m} \Rightarrow \frac{\lambda}{4}=0.8$
$\Rightarrow\lambda=3.2 \mathrm{~m} . $
$\therefore v=n \lambda=120 \times 3.2=384 \mathrm{~m} / \mathrm{s}$

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