Question

A tuning fork gives 4 beats with 50 cm length of a sonometer wire if length of the wire is shortened by 1 cm . the no . of beats still the same. The frequency of the fork is:

• A. 306
• B. 400
• C. 404
• D. 384

Answer 1

The correct option is C 404

Frequency of sonometer wire is given by:

$v=\frac{1}{21}\sqrt{\frac{T}{\mu }}$

So, for small change in $L,V$ changes by:

$=\frac{\Delta v}{v}=\frac{\Delta L}{L}$ ....... $\left(1\right)$

If at $L=50cm$ frequency of sonometer is $v_1$; it producer $4Hz$ beats with the fork (having frequency $v_o$)

$\Rightarrow (v_0-v_1=4)$ .......... $\left(2\right)$

When length is shortened $(\Delta l=-1)$ it produced frequency $v_2$

So, $\frac{v_2-v_1}{v_1}=\frac{1}{50}$ [from $\left(1\right)$]

$\Rightarrow \frac{V_2}{V_1}=\frac{51}{50}$

$V_2$ also produced $4Hz$ beat with fork.

$\Rightarrow v_2-v_0=4\Rightarrow (\frac{51}{50}{v}_1-v_0=4)$ ......... $\left(3\right)$

Solving $\left(2\right)$ & $\left(3\right)$ gives

$v_0=40Hz$