Two wires of the same material and same cross-section are stretch… - Vidyadip

Question

Two wires of the same material and same cross-section are stretched on a sonometer. One wire is loaded with $1.5\ kg$ and another is loaded with $6\ kg.$ The vibrating length of first wire is $60\ cm$ and its fundamental frequency of vibration is the same as that of the second wire. Calculate vibrating length of the other wire.

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Answer

Data $: m_1=m_2=m_{,} L_1=60 cm =0.6 m _{,} T_1=1.5 kg =14.7 N , T _2=6 kg =58.8 N$
$n_1=\frac{1}{2 L_1} \sqrt{\frac{T_1}{m}}$ and $n_2=\frac{1}{2 L_2} \sqrt{\frac{T_2}{m}}$
But, $n_1=n_2$
$\therefore \frac{1}{2 L_1} \sqrt{\frac{T_1}{m}}=\frac{1}{2 L_2} \sqrt{\frac{T_2}{m}}$
$ L_2=\sqrt{\frac{T_2}{T_1}} \times L_1$
$=\sqrt{\frac{58.8}{14.7}} \times 0.6=\sqrt{4} \times 0.6=1.2\ m$
The vibrating length of the second wire is $1.2\ m.$

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