Question

A 146-cm-long pipe is stopped at one end. Near the open end, there is a loudspeaker that is driven by an audio oscillator whose frequency can be varied from 10.0 to 4700 Hz. (Take the speed of sound to be 343 m/s.)

(a) What is the lowest frequency of the oscillator that will produce resonance within the tube?
  Hz

(b) What is the highest frequency that will produce resonance?
  Hz

(c) How many different frequencies of the oscillator will produce resonance? (Neglect the end correction.)

The lowest resonant frequency in this closed-at-one-end tube is its fundamental frequency which is related to its wavelength and

Answer 1

a)   fundamental frequency

$f_{1}=\frac{v}{4L}=\frac{343}{4\times 1.46}$$f_{1}=\frac{v}{4L}=\frac{343}{4\times 1.46}=58.73Hz$

will produce resonance with the tube

fi

B) for Highest fequency

$n\times f_{1}\leq4700Hz$ where n should be largest odd numver

$n\leq \frac{4\times 1.46\times 4700}{343}=80.023$

$n=79$

but n should be odd number

so largest frequency with resonance

$f_{h}=\frac{79\times 343}{4\times 1.46}=4639.897Hz$

C) as $2m+1=79\Rightarrow m=39$

so total frequencies with resonance =39+1=40 frequencies