JEE MAIN - Physics (2006 - No. 31)
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
$$105$$ $$Hz$$
$$1.05$$ $$Hz$$
$$1050$$ $$Hz$$
$$10.5$$ $$Hz$$
Explanation
Given $${{nv} \over {2\ell }} = 315$$ and $$\left( {n + 1} \right){v \over {2\ell }} = 420$$
$$ \Rightarrow {{n + 1} \over n} = {{420} \over {315}} \Rightarrow n = 3$$
Hence $$3 \times {v \over {2\ell }} = 315 \Rightarrow {v \over {2\ell }} = 105Hz$$
Lowest resonant frequency is when $$n=1$$
Therefore lowest resonant frequency $$ = 105\,Hz.$$
$$ \Rightarrow {{n + 1} \over n} = {{420} \over {315}} \Rightarrow n = 3$$
Hence $$3 \times {v \over {2\ell }} = 315 \Rightarrow {v \over {2\ell }} = 105Hz$$
Lowest resonant frequency is when $$n=1$$
Therefore lowest resonant frequency $$ = 105\,Hz.$$
Comments (0)
