Stationary waves

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A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is _______Hz

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Solution

$\begin{array}{l} S p e e d o f s o u n d i n a g a s = V = \sqrt{\frac{B}{ρ}} \\ ∴ \frac{V_{p i p e}}{V_{a i r}} = \frac{\sqrt{\frac{B}{2 ρ}}}{\sqrt{\frac{B}{ρ}}} = \frac{1}{\sqrt{2}} \\ \Rightarrow V_{p i p e} = \frac{V_{a i r}}{\sqrt{2}} = \frac{300}{\sqrt{2}} m / s \\ N o w, f r e q u e n c y i n o p e n o r g a n p i p e = f_{n} = \frac{(n + 1) V_{p i p e}}{2 l} \\ F u n d a m e n t a l f r e q u e n c y = f_{0} = \frac{V_{p i p e}}{2 l} \\ S e c o n d H a r m o n y = f_{1} = \frac{2 V_{p i p e}}{2 l} \\ T h u s, f r e q u e n c y d i f f e r e n c e = f_{1} - f_{0} = \frac{V_{p i p e}}{2 l} = \frac{300}{2 \sqrt{2}} = 106.38 H z (I f \sqrt{2} = 1.41) \\ o r f_{1} - f_{0} = 106.08 H z (I f \sqrt{2} = 1.414) \end{array}$

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Stationary waves

Remember concepts with our Masterclasses.

A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is _______Hz

Speed of sound in a gas =V=Bρ​∴VpipeVair=B2ρBρ=12⇒Vpipe=Vair2=3002m/s​Now, frequency in open organ pipe =fn=n+1Vpipe2lFundamental frequency =f0=Vpipe2l​Second Harmony = f1=2Vpipe2lThus, frequency difference =f1−f0=Vpipe2l=30022=106.38Hz If2=1.41or f1−f0=106.08Hz If2=1.414

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