Question

The length of the wire shown in figure between the pulley is 1.5 m and its mass is 12.0 g. 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.

• A. 140 Hz
• B. 210 Hz
• C. 70 Hz
• D. 30 Hz

Answer 1

The correct option is C

70 Hz

First let's calculate the tension in the string

Free body diagram of one of the mass

Now for two "loops” to form

we know the string is in its 2nd harmonic

$f_2=2\times \frac{v}{2L}=\frac{v}{L}$

we need to calculate velocity v for that

$v=\sqrt{\frac{T}{\mu }}and\mu =\frac{mass}{length}=\frac{12\times {10}^{-3}}{1.5}$

Even though the string has mass ,we can assume tension to be uniform because the mass of the string is

very less compared to the blocks.

$v=\sqrt{\frac{90}{\left(\frac{12\times {10}^{-3}}{1.5}\right)}}=106.06m{s}^{-1}$

$v_2=\frac{106.06}{1.5}\stackrel{\sim }{\sim }70Hz$