Two sound waves expressed in CGS units given by y 1-0-3 sin2 - vt - x and y 2-0-4 sin vt - x - and interfere- The resultantamplitude at a place where phase difference is -2 will be -MP PET 1991-A- 0-1 cmB- 1-10-7 cmC- 0-5 cmD- 0-7 cm

The correct option is C 0.5 cm Resultant ampliyude = √(a_1^2+a_2^2+2a_1a_2 cos ϕ) =√(0.3^2+0.4^2+2× 0.3× 0.4 × cos π2)=0.5 cm ...

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YDSE Problems II

Two sound wav...

Question

Two sound waves (expressed in CGS units) given by $y_{1} = 0.3 s i n \frac{2 π}{λ} (v t - x)$ and $y_{2} = 0.4 s i n (v t - x + θ)$ and interfere. The resultant amplitude at a place where phase difference is $\frac{π}{2}$ will be

[MP PET 1991]

0.7 cm

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0.1 cm

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0.5 cm

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Solution

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Similar questions

y_{1} = 0.3 s i n \frac{2 π}{λ} (v t - x)

y_{2} = 0.4 s i n \frac{2 π}{λ} (v t - x + θ)

\frac{π}{2}

Two waves are given by $y_{1} = a s i n (ω t - k x)$ and $y_{2} = a c o s (ω t - k x)$ The phase difference between the two waves is

[MP PMT 1993; SCRA 1996; CET 1998;EAMCET 1991; Orissa JEE 2002]

y_{1} = A_{1} sin (w t - k x)

y_{2} = A_{2} cos (w t - k x + π / 3)

y_{1} = 10^{- 6} sin [100 t + \frac{x}{50} + 0.5] m

y_{2} = 10^{- 6} cos [100 t + \frac{x}{50}] m

y_{1} = (2.00) sin (20 x - 32 t)

y_{2} = (2.00) sin (25 x - 40 t)

y_{1}, y_{2}

x

t

x

π

t = 2 s

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