Question

# A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected

Solution

## It is given that the total length of the tube is 20 cm and the different lengths of the tube to show resonance are 25.5 cm and 79.3 cm. The formula to calculate frequency in a closed organ pipe for n th node is, f= ( 2n−1 )v 4 l 1 (1) Here, the frequency of resonance is f, the length of the organ pipe for first resonance is l 1 and the speed of sound is v. The formula to calculate the frequency in a closed organ pipe for ( n+1 ) th node is, f= ( 2( n+1 )−1 )v 4 l 2 = ( 2n+1 )v 4 l 2 (2) Here, the frequency of resonance is f, the length of the organ pipe for the second resonance is l 2 and the speed of sound is v. The frequency obtained from equation (1) and equation (2) is the same. ( 2n−1 )v 4 l 1 = ( 2n+1 )v 4 l 2 ( 2n−1 ) ( 2n+1 ) = l 1 l 2 Substituting the values in the above equation, we get: ( 2n−1 ) ( 2n+1 ) = 25.5 79.3 ( 2n−1 ) ( 2n+1 ) = 1 3 3( 2n−1 )=( 2n+1 ) n=1 The frequency of node for resonance is for the first node. Substituting the values in equation (1), we get: 340= ( 2−1 )v 4( 25.5× 1 m 10 2  cm ) v= 340( 4 )( 25.5 ) ( 100 ) =346.8 m/s ≈347 m/s Thus, the estimated speed of the sound is 347 m/s . PhysicsPhysics Part-II - NCERTStandard XI

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