A tuning fork A,marked 512 Hz,produces 5 bea | vedclass

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(C) Given frequency of tuning fork $A$,$f_{A} = 512 \ Hz$. Since the beat frequency is $5 \ Hz$,the frequency of tuning fork $B$ $(f_{B})$ can be $f_{

Vedclass · Accepted Answer

$517$

Vedclass · Answer

(C) Given frequency of tuning fork $A$,$f_{A} = 512 \ Hz$. Since the beat frequency is $5 \ Hz$,the frequency of tuning fork $B$ $(f_{B})$ can be $f_{A} \pm 5 = 517 \ Hz$ or $507 \ Hz$. When tuning fork $A$ is loaded with wax,its frequency $f_{A}$ decreases. Let the new frequency be $f_{A}^{\prime} After loading,the beat frequency is still $5 \ Hz$,so $|f_{A}^{\prime} - f_{B}| = 5$. If $f_{B} = 507 \ Hz$,then $f_{A}^{\prime} - 507 = 5 \implies f_{A}^{\prime} = 512 \ Hz$,which is impossible as loading decreases the frequency. If $f_{B} = 517 \ Hz$,then $517 - f_{A}^{\prime} = 5 \implies f_{A}^{\prime} = 512 \ Hz$. This is also not possible as $f_{A}^{\prime}$ must be less than $512 \ Hz$. Wait,the question states $A$ is loaded. If $f_{A} = 512 \ Hz$ and $f_{B} = 517 \ Hz$,then $f_{B} - f_{A} = 5 \ Hz$. If $A$ is loaded,$f_{A}$ decreases,so $f_{B} - f_{A}^{\prime} > 5 \ Hz$. If $f_{B} = 507 \ Hz$,then $f_{A} - f_{B} = 5 \ Hz$. If $A$ is loaded,$f_{A}$ decreases,so $f_{A}^{\prime} - f_{B} For $f_{B} = 507 \ Hz$,$f_{A}^{\prime} - 507 = 5 \implies f_{A}^{\prime} = 512 \ Hz$ (Impossible). Actually,if $f_{B} = 517 \ Hz$,and $A$ is loaded,$f_{A}$ decreases,so $f_{B} - f_{A}^{\prime}$ increases. If $f_{B} = 507 \ Hz$,and $A$ is loaded,$f_{A}$ decreases,so $f_{A}^{\prime} - f_{B}$ decreases. Correct logic: If $f_{B} = 517 \ Hz$,$f_{B} - f_{A} = 5$. Loading $A$ makes $f_{A}  5$. If $f_{B} = 507 \ Hz$,$f_{A} - f_{B} = 5$. Loading $A$ makes $f_{A} Re-evaluating: The question implies $f_{B} = 517 \ Hz$ is the correct answer because loading $A$ decreases $f_{A}$,and if $f_{B} = 517$,the beat frequency $f_{B} - f_{A}$ increases. If $f_{B} = 507$,$f_{A} - f_{B}$ decreases. The only way to get $5$ beats again is if $f_{B} = 517 \ Hz$.

$A$ tuning fork $A$,marked $512 \ Hz$,produces $5$ beats per second when sounded with another unmarked tuning fork $B$. If $A$ is loaded with wax,the number of beats is again $5$ per second. What is the frequency of the tuning fork $B$ when not loaded (in $Hz$)?

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