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Question

(D) The frequency of the tuning fork is $
u_{T} = 280\, Hz$.Let the initial frequency of the sonometer string be $
u_{s}$.The beat frequency is give

Vedclass · Accepted Answer

$290$

Vedclass · Answer

(D) The frequency of the tuning fork is $
u_{T} = 280\, Hz$.Let the initial frequency of the sonometer string be $
u_{s}$.The beat frequency is given by $|
u_{T} - 
u_{s}| = 10\, Hz$.This implies $
u_{s} = 280 \pm 10$,so $
u_{s}$ is either $290\, Hz$ or $270\, Hz$.When the tension in the string increases,the frequency of the string $
u_{s}$ increases (since $
u_{s} \propto \sqrt{T}$).If $
u_{s} = 290\, Hz$,increasing the tension would make the frequency higher than $290\, Hz$,which would increase the beat frequency $|280 - 
u_{s}|$. However,we must check the condition: if $
u_{s} = 270\, Hz$,increasing the tension makes the frequency approach $280\, Hz$,which would decrease the beat frequency. If $
u_{s} = 290\, Hz$,increasing the tension makes the frequency move further away from $280\, Hz$,which increases the beat frequency.Since the beat frequency increases from $10$ to $11$,the string frequency must be $290\, Hz$.

$A$ tuning fork of frequency $280\, Hz$ produces $10$ beats per second when sounded with a vibrating sonometer string. When the tension in the string increases slightly,it produces $11$ beats per second. The original frequency of the vibrating sonometer string is ... $Hz$.

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