A string is stretched between fixed points separated by 75 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

Question

A string is stretched between fixed points separated by $75 cm$ . 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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Answer:

B

Solution:

$\frac{n}{2 ℓ} (v) = 315, \frac{( n + 1 )}{2 ℓ} v = 420$ Solving $\frac{v}{2 ℓ} = 105$

TestHub · Accepted Answer

2 ℓ n ​ ( v ) = 315 , 2 ℓ ( n + 1 ) ​ v = 420 Solving 2 ℓ v ​ = 105

Physics - Wave on String Question with Solution | TestHub

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