Question

When a tuning fork vibrates with 1.0 m or 1.05 m long wire of a sonometer, 5 beats per second are produced in each case. What will be the frequency of the tunning fork?

• A. 195
• B. 200
• C. 205
• D. 210

Answer 1

The correct option is C 205

For sonometer string, fundamental frequency, $\nu =\frac{v}{2L}$, where L is the length of the string.
Since $L_1$ is less than $L_2$, ${\nu }_1$ is greater than ${\nu }_2$.

As the strings make 5 beats with a common tuning fork
${\nu }_1-{\nu }_2=10$ ... (1)

Now,
${\nu }_1=\frac{v}{2L_1}$
${\nu }_2=\frac{v}{2L_2}$
$\frac{{\nu }_1}{{\nu }_2}=\frac{L_2}{L_1}$
${\nu }_1=1.05\times {\nu }_2$ (using $L_1=1$ m and $L_2=1.05$ m) ... (2)
Using Eq. (1) and (2), we get ${\nu }_2=200$ $Hz$
So the frequency of the tuning fork is $200+5=205Hz$