- Home
- Standard 11
- Physics
14.Waves and Sound
medium
The length of the wire shown in figure between the pulleys is $1.5\, m$ and its mass is $12.0\,g$. The frequency of vibration with which the wire vibrates in three loops forming antinode at the mid point of the wire is $(g = 9.8 \,m/s^2)$

A
$210 \,Hz$
B
$140 \,Hz$
C
$70 \,Hz$
D
None of these
Solution
$V=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{T}{m / l}}=\sqrt{\frac{90 \times 1.5}{12 \times 10^{-3}}}=106 \mathrm{\,m} / \mathrm{s}$
$n=\frac{3 V}{2 l}=\frac{3 \times 106}{2 \times 1.5}=106 \mathrm{\,Hz}$
Standard 11
Physics
Similar Questions
hard
hard