When a tuning fork of frequency 341 Hz is so | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the frequency of the first tuning fork be $n_A = 341 \ Hz$ and the frequency of the second tuning fork be $n_B$.The beat frequency is given by

Vedclass · Accepted Answer

$347$

Vedclass · Answer

(D) Let the frequency of the first tuning fork be $n_A = 341 \ Hz$ and the frequency of the second tuning fork be $n_B$.The beat frequency is given by $|n_A - n_B| = 6 \ Hz$.This implies $n_B = 341 \pm 6$,so $n_B$ could be $347 \ Hz$ or $335 \ Hz$.When the second tuning fork is loaded with wax,its frequency $n_B$ decreases $(n_B \downarrow)$.After loading,the new beat frequency is $2 \ Hz$.Case $1$: If $n_B = 335 \ Hz$,then $n_A - n_B = 6 \ Hz$. Loading $n_B$ makes it smaller,so $(n_A - n_B)$ increases. This contradicts the observation that the beat frequency decreased to $2 \ Hz$.Case $2$: If $n_B = 347 \ Hz$,then $n_B - n_A = 6 \ Hz$. Loading $n_B$ makes it smaller,so $(n_B - n_A)$ decreases. As $n_B$ decreases towards $n_A$,the beat frequency decreases. This matches the observation that the beat frequency decreased to $2 \ Hz$.Therefore,the natural frequency of the second tuning fork is $347 \ Hz$.

When a tuning fork of frequency $341 \ Hz$ is sounded with another tuning fork,six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first tuning fork,the number of beats is two per second. The natural frequency of the second tuning fork is: (in $Hz$)

Similar Questions

$50$ tuning forks are arranged in increasing order of their frequencies such that each gives $4 \, \text{beats/sec}$ with its previous tuning fork. If the frequency of the last fork is the octave of the first, then the frequency of the first tuning fork is ... $Hz$.

Two tuning forks have frequencies $450\, Hz$ and $454\, Hz$ respectively. On sounding these forks together,the time interval between successive maximum intensities will be .... $sec$

When two tuning forks are sounded together,$6$ beats per second are heard. One of the forks is in unison with $0.70 \ m$ length of a sonometer wire and another fork is in unison with $0.69 \ m$ length of the same sonometer wire. The frequencies of the two tuning forks are:

Two tuning forks of frequencies $320 \ Hz$ and $323 \ Hz$ are vibrated together. The time interval between a maximum sound and its adjacent minimum sound heard by an observer is

Two waves of wavelengths $99\, cm$ and $100\, cm$ both travelling with velocity $396\, m/s$ are made to interfere. The number of beats produced by them per second is

Vedclass Test Series

Exam Paper Generator

Online Exam Module