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Standard XII

Physics

Springs

Two vibrating...

Question

Two vibrating strings of the same material but lengths $L$ and $2 L$ have radii $2 r$ and $r$ respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length $L$ with frequency $v_{1}$ and the other with frequency $v_{2}$ . The ratio $\frac{v_{1}}{v_{2}}$ is given by

2

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Solution

The correct option is D $1$
$f = \frac{1}{2 l} \sqrt{\frac{T}{μ}} = \frac{1}{2 l} \sqrt{\frac{T}{3 (A)}} = \frac{1}{2 l r} \sqrt{\frac{T}{ϱ π}}$
$f \propto \frac{1}{l r}$
$∴ \frac{f_{1}}{f_{2}} = \frac{2 r \times l}{2 l \times r} = 1$

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Two vibrating strings are of same material but lengths l, 2l and radii 2r, r respectively are stretched under the same tension. Both are vibrating in their fundamental modes. The experimental arrangement is such that first string has both ends fixed, second string vibrates with one end fixed and other end as free.

If the frequencies of vibration of first and second wire are $n_{1}$ and $n_{2}$ , then $\frac{n_{1}}{n_{2}}$ = ________

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Two vibrating strings of the same material but lengths L and 2L have radii 2r and r respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length L with frequency v1 and the other with frequency v2 . The ratio v1v2 is given by

The correct option is D 1f=12l√Tμ=12l√T3(A)=12lr√Tϱπf∝1lr∴f1f2=2r×l2l×r=1

The correct option is D $1$
$f = \frac{1}{2 l} \sqrt{\frac{T}{μ}} = \frac{1}{2 l} \sqrt{\frac{T}{3 (A)}} = \frac{1}{2 l r} \sqrt{\frac{T}{ϱ π}}$
$f \propto \frac{1}{l r}$
$∴ \frac{f_{1}}{f_{2}} = \frac{2 r \times l}{2 l \times r} = 1$