First slide

Standing waves

Question

A wire is stretched between two rigid supports. It is observed that wire resonates at the frequencies $f_{1}, f_{2}, f_{3} and f_{4} (f_{4} > f_{3} > f_{2} > f_{1})$ forming 2,3,4 and 5 loops respectively. The ratio of any two resonance frequencies will be minimum for difference in the loops to be:

Moderate

A

1
B

2
C

3
D

zero

Solution

Let $f_{1} = \frac{mv}{2 L} = \frac{2 v}{2 L} . Similarly f_{4} = \frac{(n + 3)}{2 L} = \frac{5 v}{2 L}$
$f = \frac{f_{1}}{f_{4}}$ will be minimum $f = \frac{f_{1}}{f_{4}} = \frac{n}{n + 3}$

f will be minimum for difference in loops = 3.

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