
A tunning fork of known frequency \[256Hz\] makes 5 beats per second with the vibrating string of the piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
A. \[256+5\,Hz\]
B. \[256+2\,Hz\]
C. \[256-2\,Hz\]
D. \[256-5\,Hz\]
Answer
587.1k+ views
Hint: The difference in the frequency of the tunning fork and the string of the piano will give us the number of beats per second.
Formula used: \[{{f}_{2}}-{{f}_{1}}=n\]
Where,
\[{{f}_{1}}\] is the frequency of the tunningfork
\[{{f}_{2}}\] is the frequency of the Piano
\[n\] is the number of beats per second or the beat frequency
Complete step by step solution:
Given that, the frequency of the tunning fork, \[{{f}_{1}}=256\,\,Hz\]
And the strings of the piano 5 beats per second
Let \[{{f}_{2}}\] be the frequency of string of the piano
Therefore, \[{{f}_{2}}-{{f}_{1}}=\pm 5\]
Or, we can say that \[{{f}_{2}}=\left( 256+5 \right)Hz\,\,or\,\,\left( 256-5 \right)Hz\]
We know that \[f\propto \sqrt{Tension}\]
Now, according to the question, when the tension in the string increases, the beat frequency decreases by 2 beats per second. And this is possible when the string frequency is increasing but the beat frequency is decreasing.
So if we suppose \[{{f}_{2}}=261\,\,Hz\], then on increasing the tension, string frequency will be something more than 261 Hz, let the value be \[265\,\,Hz\]
Now this simply means that \[265\,-256>261-256\], Which is not possible as the no. of beats is decreasing.
Thus, the value of string frequency of the piano before increasing the tension will be \[251\,\,Hz\]or can be written as \[{{f}_{2}}=256-5\,\,Hz\]
The correct answer is (D).
Additional information: A tunning is basically a U-shaped instrument which is in the form of two pronged forks. It is used to produce a specific constant pitch when striked against any surface or body.
It emits a pure musical tone once the high overtone fades out.
Note:The frequency of the string of a piano is directly proportional to the square root of the tension in the string. It means more the tension in the string higher will be the frequency of the string.
Formula used: \[{{f}_{2}}-{{f}_{1}}=n\]
Where,
\[{{f}_{1}}\] is the frequency of the tunningfork
\[{{f}_{2}}\] is the frequency of the Piano
\[n\] is the number of beats per second or the beat frequency
Complete step by step solution:
Given that, the frequency of the tunning fork, \[{{f}_{1}}=256\,\,Hz\]
And the strings of the piano 5 beats per second
Let \[{{f}_{2}}\] be the frequency of string of the piano
Therefore, \[{{f}_{2}}-{{f}_{1}}=\pm 5\]
Or, we can say that \[{{f}_{2}}=\left( 256+5 \right)Hz\,\,or\,\,\left( 256-5 \right)Hz\]
We know that \[f\propto \sqrt{Tension}\]
Now, according to the question, when the tension in the string increases, the beat frequency decreases by 2 beats per second. And this is possible when the string frequency is increasing but the beat frequency is decreasing.
So if we suppose \[{{f}_{2}}=261\,\,Hz\], then on increasing the tension, string frequency will be something more than 261 Hz, let the value be \[265\,\,Hz\]
Now this simply means that \[265\,-256>261-256\], Which is not possible as the no. of beats is decreasing.
Thus, the value of string frequency of the piano before increasing the tension will be \[251\,\,Hz\]or can be written as \[{{f}_{2}}=256-5\,\,Hz\]
The correct answer is (D).
Additional information: A tunning is basically a U-shaped instrument which is in the form of two pronged forks. It is used to produce a specific constant pitch when striked against any surface or body.
It emits a pure musical tone once the high overtone fades out.
Note:The frequency of the string of a piano is directly proportional to the square root of the tension in the string. It means more the tension in the string higher will be the frequency of the string.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

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

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

