Two sound waves, travelling in same direction can be represented as andy1=(0.02mm)sin[(400π rads−1)(x330ms−1−t)] y2=(0.02mm)sin[(404π rads−1)(x330ms−1−t)] the wave superimpose.

y1=0.02sin[400π(x330−t)]y2=0.02sin[404π(x330−t)]y1+y2=2(0.02)sin[404π(x330−t)]cos[2π(x330−t)]=0.04sin(402330πx−402πt)cos(2πx330−2πt)

Two sound waves, travelling in same direction can be represented as andy1=(0.02mm)sin[(400π rads−1)(x330ms−1−t)] y2=(0.02mm)sin[(404π rads−1)(x330ms−1−t)] the wave superimpose.

y1=0.02sin[400π(x330−t)]y2=0.02sin[404π(x330−t)]y1+y2=2(0.02)sin[404π(x330−t)]cos[2π(x330−t)]=0.04sin(402330πx−402πt)cos(2πx330−2πt)

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Two sound waves, travelling in same direction can be represented as and
$y_{1} = (0.02 m m) \sin [(400 π r a d s^{- 1}) (\frac{x}{330 m s^{- 1}} - t)]$
$y_{2} = (0.02 m m) \sin [(404 π r a d s^{- 1}) (\frac{x}{330 m s^{- 1}} - t)]$ the wave superimpose.

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The time gap between two successive intensity maxima at a given point is 1 sec.

Distance between two nearest points where an intensity maximum is recorded Simultaneously is

330 m.

The time gap between two successive intensity maxima at a given point is 0.5 sec

Distance between two nearest points where an intensity maximum is recorded Simultaneously is

165 m

answer is A, C.

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Detailed Solution

$y_{1} = 0.02 \sin [400 π (\frac{x}{330} - t)]$

$y_{2} = 0.02 \sin [404 π (\frac{x}{330} - t)]$

$y_{1} + y_{2} = 2 (0.02)$

$\sin [404 π (\frac{x}{330} - t)] \cos [2 π (\frac{x}{330} - t)]$

$= 0.04 \sin (\frac{402}{330} π x - 402 π t) \cos (\frac{2 π x}{330} - 2 π t)$

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Two sound waves, travelling in same direction can be represented as andy1=(0.02mm)sin[(400π rads−1)(x330ms−1−t)] y2=(0.02mm)sin[(404π rads−1)(x330ms−1−t)] the wave superimpose.

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Two sound waves, travelling in same direction can be represented as and
$y_{1} = (0.02 m m) \sin [(400 π r a d s^{- 1}) (\frac{x}{330 m s^{- 1}} - t)]$
$y_{2} = (0.02 m m) \sin [(404 π r a d s^{- 1}) (\frac{x}{330 m s^{- 1}} - t)]$ the wave superimpose.