Answer
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Hint: When two sound waves of different frequency approach towards an observer, an alternatively soft and loud sound is heard due to the alternating constructive and destructive interference of the sound wave. This phenomenon is known as "beating" or producing beats. The beat frequency is equal to the absolute value of the difference in frequency of the two waves.
According to the question, each fork produces 5 beats per second so each frequency will be 5 more than the preceding one which forms an arithmetic progression. Find the first and last term of the A.P.
Complete step by step answer:
Let us first discuss how beats are produced.
When two sound waves of different frequency approach towards an observer, alternatively soft and loud sound is heard due to the alternating constructive and destructive interference of the sound wave. This phenomenon is known as "beating" or producing beats. The beat frequency is equal to the absolute value of the difference in frequency of the two waves.
According to the question, each fork produces 5 beats per second so each frequency will be 5 more than the preceding one which forms an arithmetic progression.
Let the frequency of the first fork be $x$ which is the first term of the A.P. The common difference of the A.P. will be equal to $5$ .
The A.P. is $x,x + 5,x + 10,.........,{x_{41}}$ where ${x_{41}}$, the last term of A.P. and the frequency of the last fork.
We know that the last term will be written as
${x_{41}} = x + \left( {41 - 1} \right) \times 5$
On simplifying we have
${x_{41}} = x + 200$ ……(i)
Given in the question that the frequency of last fork is double the frequency of first fork i.e.
${x_{41}} = 2x$
Substituting this value in equation (i) we have
$2x = x + 200$
On simplifying we have
$x = 200$
Therefore, ${x_{41}} = 2x = 2 \times 200 = 400$
So, the frequency of the first fork is $200$ and that of the last fork is $400$ .
Hence, option A is correct.
Note: Beats has numerous applications in our daily life such as they are used to tune musical instruments such as guitar and violin. They are also used in the sonometer experiment to adjust the vibrating length between the two bridges.
According to the question, each fork produces 5 beats per second so each frequency will be 5 more than the preceding one which forms an arithmetic progression. Find the first and last term of the A.P.
Complete step by step answer:
Let us first discuss how beats are produced.
When two sound waves of different frequency approach towards an observer, alternatively soft and loud sound is heard due to the alternating constructive and destructive interference of the sound wave. This phenomenon is known as "beating" or producing beats. The beat frequency is equal to the absolute value of the difference in frequency of the two waves.
According to the question, each fork produces 5 beats per second so each frequency will be 5 more than the preceding one which forms an arithmetic progression.
Let the frequency of the first fork be $x$ which is the first term of the A.P. The common difference of the A.P. will be equal to $5$ .
The A.P. is $x,x + 5,x + 10,.........,{x_{41}}$ where ${x_{41}}$, the last term of A.P. and the frequency of the last fork.
We know that the last term will be written as
${x_{41}} = x + \left( {41 - 1} \right) \times 5$
On simplifying we have
${x_{41}} = x + 200$ ……(i)
Given in the question that the frequency of last fork is double the frequency of first fork i.e.
${x_{41}} = 2x$
Substituting this value in equation (i) we have
$2x = x + 200$
On simplifying we have
$x = 200$
Therefore, ${x_{41}} = 2x = 2 \times 200 = 400$
So, the frequency of the first fork is $200$ and that of the last fork is $400$ .
Hence, option A is correct.
Note: Beats has numerous applications in our daily life such as they are used to tune musical instruments such as guitar and violin. They are also used in the sonometer experiment to adjust the vibrating length between the two bridges.
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