A uniform string resonates with a tuning fork | vedclass

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Question

(a)

Let length be $l$ .

$f=\sqrt{\frac{T}{\mu}} 	imes \frac{1}{2 l} \quad \dots (i)$

$f=\sqrt{\frac{T}{\mu}} 	imes \frac{4}{2 l} \quad \dot

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(a)

Let length be $l$ .

$f=\sqrt{\frac{T}{\mu}} 	imes \frac{1}{2 l} \quad \dots (i)$

$f=\sqrt{\frac{T}{\mu}} 	imes \frac{4}{2 l} \quad \dots (ii)$

or $f=\sqrt{\frac{T}{\mu}} 	imes \frac{4}{6 l} \quad \dots (iii)$.

Equating $(i)$ $(ii)$ and $(i)$ and $(iii)$

$\sqrt{\frac{T}{T_1}}=4$ and $\sqrt{\frac{T}{T_2}}=\frac{4}{3}$

Put $T=32 \,N$

$\frac{32}{16}=T_1 \frac{9}{16} 	imes 32=T_2$

$T_1=2 N T_2=18 \,N$

of the options on $T_1$ is right.

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 of length from one end, then to resonance with same frequency the maximum value of tension for string will be ........... $N$

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