Question

A tuning fork vibrating with a sonometer having 20 cm wire produces 5 beats/s. The beat frequency does not change, if the length of the wire is changed to 22 cm. The frequency of the tuning fork (in Hz) must be

• A. 200
• B. 210
• C. 105
• D. 125

Answer 1

The correct option is C 105

$\delta =\frac{v}{2l}$

So, ${\delta }_1=\frac{v}{2l_1}and{\delta }_2=\frac{v}{2l_2}$

$2\times beat={\delta }_1-{\delta }_2$

$2\times 5=\frac{v}{2l_1}-\frac{v}{2l_2}$

$2\times 5=v(\frac{1}{2\left(20\right)}-\frac{1}{2\left(22\right)})$

$v=\frac{2\times 5\times 2\times 20\times 22}{22-20}$ $=4400cm/s$

So, ${\delta }_1=\frac{4400}{2\left(20\right)}=110H_z$

So, $beat={\delta }_1-frequencyoffork$
$5=110-frequencyoffork$
$\Rightarrow frequencyoffork=105H_z$