A tuning fork of frequency 542 Hz is placed near the top of the pipe shown in th

Physics
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A tuning fork of frequency 542 Hz is placed near the top of the pipe shown in the Figure. The water level is lowered so that the length L slowly increases from an initial value of 20.0 cm. Determine the next two values of L that correspond to resonant modes. (Assume that the speed of sound in air is 343 m/s.)

 m (shorter length) 
 m (longer length)

Apr 1st, 2015

The wavelength times the frequency is the velocity of sound, so 
lambda = c / f 
At a frequency of 542 Hz and speed 343 m/s the wavelength is 0.633 m  (63.3cm)

If the tube is open at the top there is a maximum at the top and a node at the bottom at resonance. 
i.e. at L = lambda / 4 
The next resonance occurs at 3 lambda/4 then 5 lambda/4 etc. 

these correspond to 63.3 / 4 = 15.825 cm.................shortest

5lambda/4 = 79.125 cm..............longest 

Mar 31st, 2015

the answer was incorrect can you give it a try again thanks

Mar 31st, 2015

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