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The total length of a sonometer wire between fixed ends is $110\, cm$. Two bridges are placed to divide the length of wire in ratio $6 : 3 : 2$. The tension in the wire is $400\, N$ and the mass per unit length is $0.01\, kg/m$ . What is the minimum common frequency with Which three parts can vibrate ........... $Hz$ ?
$1100$
$1000$
$166$
$100$
Solution
Total length of sonometer wire, $l=110 \mathrm{cm}$ $=1.1 \mathrm{m}$
Length of wire is in ratio,$6: 3: 2$ i.e $60 \mathrm{cm}$, $30 \mathrm{cm}, 20 \mathrm{cm}$
Tension in the wire, $T=400 \mathrm{N}$
Mass per unit length, $\mathrm{m}=0.01 \mathrm{kg}$
Minimum common frequency $=?$
As we know,
Frequency, $v=\frac{1}{21} \sqrt{\frac{T}{m}}=\frac{1000}{11}-H z$
Similarly, $v_{1}=\frac{1000}{6} \mathrm{Hz}$
$v_{2}=\frac{1000}{3} \mathrm{Hz}$
$v_{3}=\frac{1000}{2} \mathrm{Hz}$
Hence common frequency $=1000 \mathrm{Hz}$
