Question

At tuning fork is vibrating at frequency 100 Hz. When another tuning fork is sounded simultaneously, 4 beats per second are heard. When some mass is added to the tuning fork of 100 Hz, beat frequency decreases. Find the frequency of the other tuning fork.

Answer 1

We have , $\Delta f=$ number of beats per second ,

given , $f_1=100Hz$ and let other frequency is $f_2$ ,

then , $f_1-f_2=100-f_2=4$ ,

therefore , $f_2=96$ or $104Hz$ ,

when some mass is added to the first tuning fork of frequency $100Hz$ , its frequency will decrease (let $f_1^{\prime }$) ,

then $f_1^{\prime }-f_2=x<4$ , (as given , beat frequency decreases) ,

as $f_1^{\prime }<100$ and $x<4$ , therefore $f_2=96Hz$