Question

A uniform string resonates with a tuning fork at a maximum tension of 32 N . If it is divided into two segments by placing a wedge at a distance one fourth if length from one end then to resonance with same frequency the max value of tension for string will be
( I got the two values of tension 2 and 18 please explain the part why we pick the answer 2 N in detail )

Answer 1

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Answer 2

Answer:

The maximum value of tension for string will be 2 N and 18 N

Explanation:

Given that,

Tension T =32 N

Let the total length is l of the string.

Length $l_{1}=l$  

If it is divided into two segments by placing a wedge at a distance one fourth.

Therefore,  

length $l_{2}=\dfrac{l}{4}$

The frequency is given by the the formula  

$n = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}$

$l = \dfrac{1}{2n}\sqrt{\dfrac{T}{\mu}}$

$l\propto\sqrt{T}$

Now,  

$(\dfrac{l_{1}}{l_{2}})^2=\dfrac{T_{1}}{T_{2}}$

$\dfrac{32}{T_{2}}=(\dfrac{l\times4}{l})^2$

$\dfrac{32}{T_{2}}=16$

$T_{2}=2\ N$

Now,

The tension will be at length $l_{2}=\dfrac{3l}{4}$

$(\dfrac{l_{1}}{l_{2}})^2=\dfrac{T_{1}}{T_{2}}$

$\dfrac{32}{T_{2}}=(\dfrac{l\times4}{3l})^2$

$\dfrac{32}{T_{2}}=\dfrac{16}{9}$

$T_{2}= 18\ N$

Hence, The maximum value of tension for string will be 2 N and 18 N.