(B) In the condition of resonance,the frequency of the alternating current $(n)$ is equal to the fundamental frequency of the vibrating wire.The funda

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(B) In the condition of resonance,the frequency of the alternating current $(n)$ is equal to the fundamental frequency of the vibrating wire.The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $l$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.Given:Length $l = 1 \, m$Tension $T = 10 \, kg \text{-} wt = 10 \times 9.8 \, N = 98 \, N$Linear mass density $\mu = 9.8 \, g/m = 9.8 \times 10^{-3} \, kg/m$Substituting the values:$n = \frac{1}{2 \times 1} \sqrt{\frac{98}{9.8 \times 10^{-3}}}$$n = \frac{1}{2} \sqrt{10000}$$n = \frac{1}{2} \times 100 = 50 \, Hz$.

Class 11 Physics Waves and Sound Transverse Stationary Waves and Sonometer

Medium

$A$ metal wire of linear mass density of $9.8 \, g/m$ is stretched with a tension of $10 \, kg$ weight between two rigid supports $1 \, m$ apart. The wire passes at its middle point between the poles of a permanent magnet,and it vibrates in resonance when carrying an alternating current of frequency $n$. The frequency $n$ of the alternating source is ..... $Hz$.

AIEEE 2003 All AIEEE

A
$25$
B
$50$
C
$100$
D
$200$

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Waves and Sound

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Transverse Stationary Waves and Sonometer

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Two uniform stretched strings $A$ and $B$,made of steel,are vibrating under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$,the ratio of the lengths of the strings is

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If the length of a stretched string is shortened by $40\%$ and the tension is increased by $44\%$,then the ratio of the final and initial fundamental frequencies is

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