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A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is :
- a)4
- b)6
- c)8
- d)12
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A tuning fork of frequency 392 Hz, resonates with 50 cm length of a st...
Given data:
Frequency of tuning fork, f = 392 Hz
Length of the string under tension, L = 50 cm
Change in length of the string, ΔL = -2%
Number of beats heard, n = ?
Constant tension, T
Formula used:
The frequency of a vibrating string is given by the formula:
f = (1/2L) √(T/m)
Where, T = tension in the string
m = mass per unit length of the string
Calculation:
Let the mass per unit length of the string be m.
Then, the frequency of the string when it is in resonance with the tuning fork is:
f = 392 Hz
Using the formula for frequency of a vibrating string, we have:
f = (1/2L) √(T/m)
⇒ T/m = (4L²f²)
Now, when the length of the string is decreased by 2%, the new length of the string becomes:
L' = L - ΔL/100 x L
⇒ L' = 50 - 2/100 x 50
⇒ L' = 49 cm
The new frequency of the string is:
f' = (1/2L') √(T/m)
⇒ f' = (1/2 x 49) √(T/m)
⇒ f' = 1/98 √(T/m)
The beat frequency is given by the difference between the frequencies of the tuning fork and the string:
n = | f' - f |
⇒ n = | 1/98 √(T/m) - 392 |
Now, since the tension in the string is constant, we can equate T/m from the two equations above:
T/m = (4L²f²) = (4L'²f'²)
Substituting the value of f' in the above equation, we get:
T/m = (4 x 49²)/(98² x 392²)
⇒ T/m = 0.000000191
Substituting the value of T/m in the equation for n, we get:
n = | 1/98 √(0.000000191) - 392 |
⇒ n = 7.84
Rounding off to the nearest integer, we get:
n = 8
Therefore, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is 8.
Frequency of tuning fork, f = 392 Hz
Length of the string under tension, L = 50 cm
Change in length of the string, ΔL = -2%
Number of beats heard, n = ?
Constant tension, T
Formula used:
The frequency of a vibrating string is given by the formula:
f = (1/2L) √(T/m)
Where, T = tension in the string
m = mass per unit length of the string
Calculation:
Let the mass per unit length of the string be m.
Then, the frequency of the string when it is in resonance with the tuning fork is:
f = 392 Hz
Using the formula for frequency of a vibrating string, we have:
f = (1/2L) √(T/m)
⇒ T/m = (4L²f²)
Now, when the length of the string is decreased by 2%, the new length of the string becomes:
L' = L - ΔL/100 x L
⇒ L' = 50 - 2/100 x 50
⇒ L' = 49 cm
The new frequency of the string is:
f' = (1/2L') √(T/m)
⇒ f' = (1/2 x 49) √(T/m)
⇒ f' = 1/98 √(T/m)
The beat frequency is given by the difference between the frequencies of the tuning fork and the string:
n = | f' - f |
⇒ n = | 1/98 √(T/m) - 392 |
Now, since the tension in the string is constant, we can equate T/m from the two equations above:
T/m = (4L²f²) = (4L'²f'²)
Substituting the value of f' in the above equation, we get:
T/m = (4 x 49²)/(98² x 392²)
⇒ T/m = 0.000000191
Substituting the value of T/m in the equation for n, we get:
n = | 1/98 √(0.000000191) - 392 |
⇒ n = 7.84
Rounding off to the nearest integer, we get:
n = 8
Therefore, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is 8.
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Community Answer
A tuning fork of frequency 392 Hz, resonates with 50 cm length of a st...
The frequency of tuning fork, f = 392 Hz.
Also
After decreasing the length by 2%, we have
From above equations,
f' = 400 Hz.
∴ Beats frequency=8 Hz.
Also
After decreasing the length by 2%, we have
From above equations,
f' = 400 Hz.
∴ Beats frequency=8 Hz.
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A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is :a)4b)6c)8d)12Correct answer is option 'C'. Can you explain this answer?
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A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is :a)4b)6c)8d)12Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is :a)4b)6c)8d)12Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is :a)4b)6c)8d)12Correct answer is option 'C'. Can you explain this answer?.
A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is :a)4b)6c)8d)12Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is :a)4b)6c)8d)12Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is :a)4b)6c)8d)12Correct answer is option 'C'. Can you explain this answer?.
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