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Standard XII

Physics

Resonance in Sound Waves

An open pipe ...

Question

An open pipe resonates with a tuning fork of frequency 500 Hz. It is observed that two successive notes are formed at distances 16 cm and 46 cm from the open end. The speed of sound in air in the pipe is

230 m/s

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300 m/s

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320 m/s

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360 m/s

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Solution

The correct option is A 300 m/s
Given: $\frac{λ}{2} = (46 - 16) c m = 30 c m$

$\Rightarrow λ = 60 c m$

And we know that frequency $= \frac{s p e e d o f s o u n d (ν)}{λ}$

$500 \times \frac{60}{100} = s p e e d o f s o u n d (ν)$

$\Rightarrow ν = 300 m / s .$

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An open pipe resonates with a tuning fork of frequency 500 Hz. It is observed that two successive notes are formed at distances 16 cm and 46 cm from the open end. The speed of sound in air in the pipe is

The correct option is A 300 m/sGiven: λ2=(46−16)cm=30cm⇒λ=60cmAnd we know that frequency =speed of sound(ν)λ500×60100=speed of sound (ν)⇒ν=300 m/s.

The correct option is A 300 m/s
Given: $\frac{λ}{2} = (46 - 16) c m = 30 c m$

$\Rightarrow λ = 60 c m$

And we know that frequency $= \frac{s p e e d o f s o u n d (ν)}{λ}$

$500 \times \frac{60}{100} = s p e e d o f s o u n d (ν)$

$\Rightarrow ν = 300 m / s .$