Question

A thin wire of mass per unit length 0.03 kg/m is rigidly clamped at points A and B and a block of weight 75 N is suspended from it as shown The
minimum frequency of a tuning fork with which the segment AB of the wire can resonate is 150 Hz. The length AB (in cm) is.
B​

Answer 1

Given:

A thin wire of mass per unit length 0.03 kg/m is rigidly clamped at points A and B and a block of weight 75 N is suspended from it as shown.

To find:

Length of AB ?

Calculation:

First of all, let's calculate the Resonance Frequency in terms of length :

For fundamental conditions:

$\therefore \: \dfrac{ \lambda}{2} = L$

$\implies \: \lambda = 2L$

Now, frequency be f :

$\therefore \: f = \dfrac{v}{ \lambda}$

$\implies\: f = \dfrac{v}{2L}$

Now, velocity can be written in form of TENSION and LINEAR MASS DENSITY:

$\implies\: f = \dfrac{ \sqrt{ \frac{T}{ \mu} } }{2L}$

$\implies\: L=\dfrac{ \sqrt{ \frac{T}{ \mu} } }{2f}$

$\implies\: L=\dfrac{ \sqrt{ \frac{75}{ 0.03} } }{2 \times 150}$

$\implies\: L=\dfrac{ \sqrt{2500}}{2 \times 150}$

$\implies\: L=\dfrac{ 50}{2 \times 150}$

$\implies\: L=\dfrac{ 1}{6}$

$\implies\: L=0.1667 \: m$

$\implies\: L=16.67 \: cm$

So, length of AB segment is 16.67 cm.

Answer 2

Length of AB Segment is 166.67 cm .

Hope this will help u