A 130-cm-long pipe is stopped at one end. Near the open end, there is a loudspea

Question

A 130-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 5500 Hz. (Take the speed of sound to be 343 m/ s.) What is the lowest frequency of the oscillator that will produce resonance within the tube? Hz What is the highest frequency that will produce resonance? Hz 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 wave speed. You can use the relationship between the nth harmonic and the fundamental frequency to find the highest frequency produced by the oscillator that will result in resonance.

Explanation / Answer

standing waves are formed with wavelegth satisfying the relation

(L=(n+rac{1}{2])rac{lambda}{2})

where n is an integer greater than or equal to 0

so, frequency will be given by:

( u = rac{v}{2L}(n+rac{1}{2}))

now, v/2L = 343/(2*1.30) = 132 Hz

So, ( u = 66(2n+1))

Thus, lowest frequency is n=0 which is 66 hz

since max frequency of audio oscillator is 5500 hz,

so max frequency will satisfy

66(2n+1)<=5500

2n+1= 83.33 = 83

so, n=41

Thus, highest frequency possible is

83*66= 5478 Hz

number of different modes = n+1 = 42

we add 1 to n as n=0 is also a mode.

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