A tuning fork of frequency $$280Hz$$ produces $$10$$ beats per $$sec$$ when sounded with a vibrating sonometer string (in$$ fundamental $$mode). When the tension in the string increases slightly, it produces $$11$$ beats per $$sec$$. The original frequency of the vibrating sonometer string in Hz is:

Question

A tuning fork of frequency $$280Hz$$ produces $$10$$ beats per $$sec$$ when sounded with a vibrating sonometer string (in$$ fundamental $$mode).  When the tension in the string increases slightly, it produces $$11$$ beats per $$sec$$.  The original frequency of the vibrating sonometer string in Hz is:

Infinity Learn · Accepted Answer

Given frequency of tunning fork  ϑ$$=280$$ HzLet the original frequency of the sonometer is ϑsSince number of beats produced is $$10$$⇒$$280$$−ϑ$$S=10$$⇒ϑ$$S=280+10$$∴ϑ$$S=290HzThe$$ frequency of the wire will be $$290$$ Hz as its frequuency is increasing with increase in tension and producing more number of beats.

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A tuning fork of frequency 280Hz produces 10 beats per sec when sounded with a vibrating sonometer string (in fundamental mode). When the tension in the string increases slightly, it produces 11 beats per sec. The original frequency of the vibrating sonometer string in Hz is:

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