Question 216: Waves

Question

A string is stretched between fixed points separated by $$75.0$$ $$cm.$$ It is observed to have resonant frequencies of $$420$$ $$Hz$$ and $$315$$ $$Hz$$. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

Accepted Answer

Given $${{nv} \over {2\ell }} = 315$$ and $$\left( {n + 1} ight){v \over {2\ell }} = 420$$
$$ \Rightarrow {{n + 1} \over n} = {{420} \over {315}} \Rightarrow n = 3$$
Hence $$3 	imes {v \over {2\ell }} = 315 \Rightarrow {v \over {2\ell }} = 105Hz$$
Lowest resonant frequency is when $$n=1$$
Therefore lowest resonant frequency $$ = 105\,Hz.$$

Answer

$$105$$ $$Hz$$

Answer

$$1.05$$ $$Hz$$

Answer

$$1050$$ $$Hz$$

Answer

$$10.5$$ $$Hz$$

Waves

A string is stretched between fixed points separated by $75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz$. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

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