Question #117373
A pipe, open at both ends, is sounding its fundamental note. calculate the frequency of the note if the pipe is 56 cm long. Neglect end corrections of the pipe. Take the speed of sound in air as 350 m/s.
A tuning fork vibrating at a frequency of 512 Hz is held on top of a jar filled with water and fitted with a tap at the bottom. If the jar is 60 cm high and the speed of sound is 350 m/s, describe what occurs as the jar slowly emptied.
1
Expert's answer
2020-05-21T13:59:02-0400

1)

f=v2l=3502(0.56)=312.5 Hzf=\frac{v}{2l}=\frac{350}{2(0.56)}=312.5\ Hz

2)


λ=vf=350512=0.684 m\lambda=\frac{v}{f}=\frac{350}{512}=0.684\ m

1st resonance will be at


140.684=0.17 m=17 cm\frac{1}{4}0.684=0.17\ m=17\ cm

2nd resonance at

340.684=0.51 m=51 cm\frac{3}{4}0.684=0.51\ m=51\ cm


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