Sound waves of frequency $660 Hz$ fall normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particle has maximum amplitude of vibration is (velocity of sound in air is $330 m/s)$

0.125 m

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0.5 m

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0.125 m

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2 m

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Solution

The correct option is A $0.125 m$

$λ = \frac{v}{f} = \frac{330}{660} = \frac{1}{2}$
Amplitude is maximum at
$\Rightarrow \frac{λ}{4} = \frac{1}{8}$
$\Rightarrow 0.125 m$

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Sound waves of frequency 660 Hz fall normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particle has maximum amplitude of vibration is (velocity of sound in air is 330 m/s)

The correct option is A 0.125 m λ=vf=330660=12 Amplitude is maximum at ⇒λ4=18 ⇒0.125 m

Sound waves of frequency $660 Hz$ fall normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particle has maximum amplitude of vibration is (velocity of sound in air is $330 m/s)$

The correct option is A $0.125 m$

$λ = \frac{v}{f} = \frac{330}{660} = \frac{1}{2}$
Amplitude is maximum at
$\Rightarrow \frac{λ}{4} = \frac{1}{8}$
$\Rightarrow 0.125 m$