Question

A tuning fork of fork of frequency 256 Hz produces 4 beats per second with a wire of length 25 cm vibrating in its fundamental mode. The beat frequency decreases when the length is slightly shortened. The minimum length by which the wire be shortened so that it produces no beats with the tuning fork is $40\times {10}^{-x}cm$. Find $x$.

Answer 1

Let $n$ be the fundamental frequency of the string of length $25cm$ .

given , frequency of fork $=256Hz$ ,

beat frequency $=4$ beats/ second ,

therefore , $n\pm 256=4$ ,

or $n=4\pm 256=260$ or $252Hz$ ,

when the string is shortened , its fundamental frequency increases due to law of length and it is given that beat frequency is decreasing on shortening the string's length .

If fundamental frequency of string is 260Hz , then beat frequency will increase when length is shortened (string frequency is increased) but it is not happening therefore string's frequency is 252Hz .

Now , after shortening the length of string , there are no beats it means , fundamental frequency of string is now 256 Hz ,

we have , initial length $l=25cm=0$ ,

length after shortening $=l^{\prime }\left(let\right)$ ,

initial frequency $n=252Hz$ ,

final frequency $=256Hz$ ,

therefore , by law of length ,

$n\propto 1/l$ ,

$l^{\prime }/l=252/256$ ,

or $l^{\prime }=l\left(252/256\right)=25\left(252/256\right)=24.60cm$ ,

hence , $l^{\prime }{l}^{\prime }=25-24.60=0.4cm$