Question

A tuning fork vibrating with a sonometer in fundamental mode having a wire of length $20\text{cm}$ produces $5$ beats per second. The beat frequency does not change if the length of the wire is changed to $21\text{cm}$. The frequency of the tuning fork is

• A. $200\text{Hz}$
• B. $210\text{Hz}$
• C. $205\text{Hz}$
• D. $215\text{Hz}$

Answer 1

The correct option is C $205\text{Hz}$

Let the frequency of the tuning fork is
$f$.
The frequency of vibrations of sonometer wire is,
$f=\frac{nv}{2l}$ [two ends fixed]
Fundamental frequency$(n=1)$,
$f_o=\frac{v}{2l}$
Now, beat frequency,
$f_b=|f_o-f|=|\frac{v}{2l}-f|$
When $l=20\text{cm}=0.20\text{m}$
$f_b=\frac{v}{2\times 0.20}-f$
$f_b=\frac{v}{2\times 0.20}-f=5\text{Hz}$ .......(1) [given]
When $l=21\text{cm}=0.21\text{m}$
$f_b=f-\frac{v}{2\times 0.21}$
[frequency of sonometer decreases as length increases]
So,
$\frac{v}{2\times 0.20}-f=f-\frac{v}{2\times 0.21}$
[given]
$\Rightarrow 2f=\frac{v}{0.40}+\frac{v}{0.42}$ .........(2)
$\Rightarrow 2(\frac{v}{0.40}-5)=\frac{v}{0.40}+\frac{v}{0.42}$
[from (1)]
On solving we get, $v=84\text{m/s}$
Putting this in (2),
$2f=\frac{84}{0.40}+\frac{84}{0.42}$
$\Rightarrow f=205\text{Hz}$