A loudspeaker that produces signals from $$50$$ to $$500$$ Hz is placed at the open end of a closed tube of length $$1.1$$ m. The lowest and the highest frequency that excites resonance in the tube are fl and fh, respectively. The velocity of sound is $$330$$ $$m/s$$. Then

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Infinity Learn · Accepted Answer

For a closed $$tubefn=nv4L$$, where n $$=$$ $$1$$,$$3$$,$$5$$,$$7$$,...$$(odd$$ $$numbers)L=1.1m$$,$$v=330m/sfn=n$$×$$3304$$×$$1.1=500Hz$$⇒$$n=6.66$$ S maximum value of n is $$5$$ and minimum value is $$1$$. Hence fl $$=$$ fmin $$=$$ $$1$$×$$3304$$×$$1.1$$ Hz $$=$$ $$75$$ Hzand fh  $$=fmax=$$ $$5$$×$$3304$$×$$1.1$$ m $$=$$ $$375$$ Hz

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