Phase difference between two points separated by 0.8 m in a wave of frequency 120 Hz is {90^o} . The

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14.Waves and Sound

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The phase difference between two points separated by $0.8 m$ in a wave of frequency $120 Hz$ is ${90^o}$. Then the velocity of wave will be ............ $\mathrm{m/s}$

A

$192$

B

$360 $

C

$710 $

D

$384$

Solution

(d) Path difference $\Delta = \frac{\lambda }{{2\pi }} \times \phi = \frac{\lambda }{{2\pi }} \times \frac{\pi }{2} = \frac{\lambda }{4}$

$\Delta = 0.8 m $ ==> $\frac{\lambda }{4} = 0.8\,$

==> $\lambda = 3.2 m.$

$ v = n \lambda = 120 × 3.2 = 384 m/s$

Standard 11

Physics

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