
Two tuning forks P and Q sounded together and 6 beats per second are heard. P is in unison with a 30 cm air column open at both ends and Q is in resonance when the length of the air column is increased by 2 cm. The frequencies of forks P and Q are
A.) 90 Hz and 84 Hz
B.) 100 Hz and 106 Hz
C.) 96 Hz and 90 Hz
D.) 206 Hz and 200 Hz
Answer
537k+ views
Hint: We will start solving this question by finding the frequencies of the two tuning forks given, then their difference will give us the number of beats per second. Upon equating the difference in frequencies with the given number of beats, we will get the velocity of both tuning forks and with which we can find their actual frequencies.
Formula used:
Frequency of a tuning fork, $f=\dfrac{v}{2l}$, where v is the velocity of sound wave generated by the tuning fork and l is the length of air column tuning fork .
Complete step by step solution:
We have been given two tuning forks P and Q who when sounded together, 6 beats per second is heard.
Length of the air column of the tuning fork P is 0.30 m and it starts resonating with Q when tuning fork Q has 0.32 m of air column.
We know that frequency of a tuning fork, $f=\dfrac{v}{2l}$, where v is the velocity of sound wave generated by the tuning fork and l is the length of air column tuning fork is in unison with.
Therefore, frequency of tuning fork P and Q can be given by ${f}_{P}=\dfrac{v}{2\times 0.30}\; Hz$ and ${f}_{Q}=\dfrac{v}{2\times 0.32}\; Hz$
According to question, upon resonance 6 beats per second is heard means ${f}_{P}-{f}_{Q}=6$
$\implies \dfrac{v}{2\times 0.30}-\dfrac{v}{2\times 0.32}=6$
$\implies v=\dfrac{2\times 0.30\times 0.32\times 6}{0.32-0.30}=57.6\; ms^{-1}$
Thus, frequency of tuning fork P, ${f}_{P}=\dfrac{57.6}{2\times 0.30}=96\; Hz$
And, frequency of tuning fork Q, ${f}_{Q}=\dfrac{57.6}{2\times 0.32}=90\; Hz$
Hence, option c is the correct answer.
Note: Two different waveforms are generated when the tuning forks start to sound together. Betas are the interference points between the two tuning forks. When the frequencies of the two tuning forks are the same, the waves show a doubled amplitude.
Formula used:
Frequency of a tuning fork, $f=\dfrac{v}{2l}$, where v is the velocity of sound wave generated by the tuning fork and l is the length of air column tuning fork .
Complete step by step solution:
We have been given two tuning forks P and Q who when sounded together, 6 beats per second is heard.
Length of the air column of the tuning fork P is 0.30 m and it starts resonating with Q when tuning fork Q has 0.32 m of air column.
We know that frequency of a tuning fork, $f=\dfrac{v}{2l}$, where v is the velocity of sound wave generated by the tuning fork and l is the length of air column tuning fork is in unison with.
Therefore, frequency of tuning fork P and Q can be given by ${f}_{P}=\dfrac{v}{2\times 0.30}\; Hz$ and ${f}_{Q}=\dfrac{v}{2\times 0.32}\; Hz$
According to question, upon resonance 6 beats per second is heard means ${f}_{P}-{f}_{Q}=6$
$\implies \dfrac{v}{2\times 0.30}-\dfrac{v}{2\times 0.32}=6$
$\implies v=\dfrac{2\times 0.30\times 0.32\times 6}{0.32-0.30}=57.6\; ms^{-1}$
Thus, frequency of tuning fork P, ${f}_{P}=\dfrac{57.6}{2\times 0.30}=96\; Hz$
And, frequency of tuning fork Q, ${f}_{Q}=\dfrac{57.6}{2\times 0.32}=90\; Hz$
Hence, option c is the correct answer.
Note: Two different waveforms are generated when the tuning forks start to sound together. Betas are the interference points between the two tuning forks. When the frequencies of the two tuning forks are the same, the waves show a doubled amplitude.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE
