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Two identical strings of the same material, same diameter and same length are in unison when stretched by the same tension. If the tension on one string is increased by 21%, the number of beats heard per second is 10. The frequency of note in Hertz when the strings are in unison is
- a)210
- b)200
- c)110
- d)100
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Two identical strings of the same material, same diameter and same len...
For transverse vibration in the string, frequency is given by n=1/2l x (T/m)^1/2(where T is tension, m is the mass per unit length)
Hence n1 = 1/2l x (T1/m)^1/2 ---(i) and
n2=1/2l x (T2/m)^1/2 ----(ii)
Number of beat n2 − n1=10 -----(iii)
Given T2 = T1 + (21/100) x T1=1.21 T1 ---(iv)
Putting the value of T2 in Eq. (ii), we get n2=1/2l x [1.21 x T1/m]^1/2 = 1.1 x 1/2l x (T1/m)^1/2
n2 = 1.1 x n1 ---(v)
Putting the value of n2 from Eq. (v) in Eq. (iii), we get, 1.1n1 − n1=10
0.1 x n1 = 10 n1 = 100Hz
Most Upvoted Answer
Two identical strings of the same material, same diameter and same len...
Given:
- Two identical strings of the same material, same diameter, and same length
- The tension on one string is increased by 21%
- The number of beats heard per second is 10
To find:
The frequency of the note in Hertz when the strings are in unison
Explanation:
When two strings are in unison, it means they vibrate at the same frequency. In this case, let's assume the frequency of the note produced by the strings is 'f' Hertz.
When the tension on one string is increased by 21%, the frequency of that string also changes. Let's assume the new frequency of the string with increased tension is 'f1' Hertz.
When two strings with slightly different frequencies are played together, they produce a phenomenon called beats. The number of beats heard per second is equal to the difference in frequencies between the two strings.
In this case, the number of beats heard per second is given as 10. So, we can write the equation as:
Number of beats = |f - f1| = 10
Since the strings are identical and have the same material, diameter, and length, the only difference in their frequencies is due to the change in tension.
Let's assume the tension on the first string is 'T' and the tension on the second string is 'T1' (after increasing it by 21%).
Now, we can write the equation for the change in frequency as:
Δf = f1 - f = k(T1 - T)
Calculating the change in tension:
Let's assume the initial tension on both strings is 'T0'.
The tension on the first string is increased by 21%, so the new tension (T1) is:
T1 = T0 + 0.21T0 = 1.21T0
Substituting the values:
Δf = f1 - f = k(T1 - T)
Δf = f1 - f = k(1.21T0 - T0)
Δf = f1 - f = k(0.21T0)
Δf = k(0.21T0)
Since the tension on one string is increased, the frequency of that string increases. So, we have:
f1 = f + Δf
Substituting the value of Δf:
f1 = f + k(0.21T0)
Calculating the number of beats:
Number of beats = |f - f1| = 10
Substituting the values:
|f - (f + k(0.21T0))| = 10
|k(0.21T0)| = 10
Since the number of beats is a positive value, we can ignore the absolute value sign:
k(0.21T0) = 10
Simplifying the equation:
k = 10 / (0.21T0)
k = 47.62 / T0
Calculating the frequency of the note:
Substituting the value of k in the equation f1 = f + k(0.21T0):
f1 = f + (47.62 / T0)(0.21T0)
f1 = f + 10
Since
- Two identical strings of the same material, same diameter, and same length
- The tension on one string is increased by 21%
- The number of beats heard per second is 10
To find:
The frequency of the note in Hertz when the strings are in unison
Explanation:
When two strings are in unison, it means they vibrate at the same frequency. In this case, let's assume the frequency of the note produced by the strings is 'f' Hertz.
When the tension on one string is increased by 21%, the frequency of that string also changes. Let's assume the new frequency of the string with increased tension is 'f1' Hertz.
When two strings with slightly different frequencies are played together, they produce a phenomenon called beats. The number of beats heard per second is equal to the difference in frequencies between the two strings.
In this case, the number of beats heard per second is given as 10. So, we can write the equation as:
Number of beats = |f - f1| = 10
Since the strings are identical and have the same material, diameter, and length, the only difference in their frequencies is due to the change in tension.
Let's assume the tension on the first string is 'T' and the tension on the second string is 'T1' (after increasing it by 21%).
Now, we can write the equation for the change in frequency as:
Δf = f1 - f = k(T1 - T)
Calculating the change in tension:
Let's assume the initial tension on both strings is 'T0'.
The tension on the first string is increased by 21%, so the new tension (T1) is:
T1 = T0 + 0.21T0 = 1.21T0
Substituting the values:
Δf = f1 - f = k(T1 - T)
Δf = f1 - f = k(1.21T0 - T0)
Δf = f1 - f = k(0.21T0)
Δf = k(0.21T0)
Since the tension on one string is increased, the frequency of that string increases. So, we have:
f1 = f + Δf
Substituting the value of Δf:
f1 = f + k(0.21T0)
Calculating the number of beats:
Number of beats = |f - f1| = 10
Substituting the values:
|f - (f + k(0.21T0))| = 10
|k(0.21T0)| = 10
Since the number of beats is a positive value, we can ignore the absolute value sign:
k(0.21T0) = 10
Simplifying the equation:
k = 10 / (0.21T0)
k = 47.62 / T0
Calculating the frequency of the note:
Substituting the value of k in the equation f1 = f + k(0.21T0):
f1 = f + (47.62 / T0)(0.21T0)
f1 = f + 10
Since
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Two identical strings of the same material, same diameter and same length are in unison when stretched by the same tension. If the tension on one string is increased by 21%, the number of beats heard per second is 10. The frequency of note in Hertz when the strings are in unison isa) 210 b) 200 c) 110 d) 100 Correct answer is option 'D'. Can you explain this answer?
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Two identical strings of the same material, same diameter and same length are in unison when stretched by the same tension. If the tension on one string is increased by 21%, the number of beats heard per second is 10. The frequency of note in Hertz when the strings are in unison isa) 210 b) 200 c) 110 d) 100 Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two identical strings of the same material, same diameter and same length are in unison when stretched by the same tension. If the tension on one string is increased by 21%, the number of beats heard per second is 10. The frequency of note in Hertz when the strings are in unison isa) 210 b) 200 c) 110 d) 100 Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two identical strings of the same material, same diameter and same length are in unison when stretched by the same tension. If the tension on one string is increased by 21%, the number of beats heard per second is 10. The frequency of note in Hertz when the strings are in unison isa) 210 b) 200 c) 110 d) 100 Correct answer is option 'D'. Can you explain this answer?.
Two identical strings of the same material, same diameter and same length are in unison when stretched by the same tension. If the tension on one string is increased by 21%, the number of beats heard per second is 10. The frequency of note in Hertz when the strings are in unison isa) 210 b) 200 c) 110 d) 100 Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two identical strings of the same material, same diameter and same length are in unison when stretched by the same tension. If the tension on one string is increased by 21%, the number of beats heard per second is 10. The frequency of note in Hertz when the strings are in unison isa) 210 b) 200 c) 110 d) 100 Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two identical strings of the same material, same diameter and same length are in unison when stretched by the same tension. If the tension on one string is increased by 21%, the number of beats heard per second is 10. The frequency of note in Hertz when the strings are in unison isa) 210 b) 200 c) 110 d) 100 Correct answer is option 'D'. Can you explain this answer?.
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