Question

The length of the wire between the pulleys is $1.5\text{m}$ and its mass is $12\text{gm}$. Find the frequency of vibration, with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest.
Use value of $\sqrt{11250}\approx 106$

• A. $35.6\text{Hz}$
• B. $40.3\text{Hz}$
• C. $70.6\text{Hz}$
• D. $80.7\text{Hz}$

Answer 1

The correct option is C $70.6\text{Hz}$


The wire between pulley will vibrate in 2 loops.
The wire in contact with pulley will behave as fixed point.
$\Rightarrow$ for two loops, $n=2$
$L=n\times \left(\frac{\lambda }{2}\right)$
$\Rightarrow 1.5=2\times \frac{\lambda }{2}$
$\therefore \lambda =1.5\text{m}$
Mass per unit length
$\mu =\frac{m}{L}=\frac{12\times {10}^{-3}}{1.5}\text{kg/m}$
From equilibrium condition of block.
$T=mg=90N$
Velocity of a wave on a stretched string
$\Rightarrow v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{90\times 1.5}{12\times {10}^{-3}}}$
$\Rightarrow v=\sqrt{11250}\approx 106$
Applying, $v=f\lambda$
$\Rightarrow f=\frac{v}{\lambda }=\frac{106}{\lambda }=\frac{106}{1.5}=70.6\text{Hz}$