A tuning fork gives $$5$$ beats with another tuning fork of frequency $$100$$ Hz. When the first tuning fork is loaded with wax, then the number of beats remains unchanged, then what will be the frequency of the first tuning fork

Question

Infinity Learn · Accepted Answer

Suppose nA $$=$$ known frequency $$=$$ $$100$$ Hz, nB $$=$$ ? x $$=$$ $$5$$ bps, which remains unchanged after loadingUnknown tuning fork is loaded so nB↓Hence nA – nB ↓ $$=$$ x       ... (i)             nB ↓ – nA $$=$$ x       ... (ii) From equation (i), it is clear that as nB decreases, beat frequency. (i$$.e. nA – $$(nB)new) can never be x again.From equation (ii), as nB↓, beat frequency (i$$.e. $$(nB)new$$ – $$nA) decreases as long as $$(nB)new$$ remains greater than nA, If $$(nB)new$$ become lesser than nA the beat frequency will increase again and will be x. Hence this is correct.So, nB $$=$$ nA $$+$$ x $$=$$ $$100$$ $$+$$ $$5$$ $$=$$ $$105$$ Hz.

A tuning fork gives 5 beats with another tuning fork of frequency 100 Hz. When the first tuning fork is loaded with wax, then the number of beats remains unchanged, then what will be the frequency of the first tuning fork

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material