Question

A tuning fork when vibrating along with a sonometer wire produces 5 beats per second when the length of the wire is either 25 cm or 26 cm. The frequency of tuning fork is

Answer 1

255 Hz

Answer 2

For a sonometer wire, the fundamental frequency is given by

$$f = \frac{k}{L},$$

with $k$ constant. Thus,

$$f_{25}=\frac{k}{25},\quad f_{26}=\frac{k}{26}.$$

Since the tuning fork produces 5 beats per second in each case, and the wire’s frequency decreases with increasing length, the tuning fork frequency $f_t$ must lie between these two frequencies. Therefore,

$$f_{25}-f_t=5 \quad \text{and} \quad f_t-f_{26}=5.$$

Adding:

$$f_{25}-f_{26}=10.$$

Substitute:

$$\frac{k}{25}-\frac{k}{26} = k\left(\frac{26-25}{25\times26}\right) = \frac{k}{650} = 10.$$

Thus,

$$k=10\times 650=6500.$$

Now, using $f_{25}-f_t=5$:

$$f_t = f_{25}-5 = \frac{6500}{25}-5=260-5=255\,\text{Hz}.$$