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The fundamental frequency of the sonometer wire Increases by 9 Hz, if its tension is increased by 69%, keeping the length constant. The frequency of the wire is
- a)42 Hz
- b)24 Hz
- c)30 Hz
- d)36 Hz
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The fundamental frequency of the sonometer wire Increases by 9 Hz, if...
We know that, frequency of vibration of a hed string is given by
View all questions of this test
where, T = tension in string
m = mass/length
and l = length of string.
New frequency,
⇒ 13v = 10v + 90
v = 30Hz
Most Upvoted Answer
The fundamental frequency of the sonometer wire Increases by 9 Hz, if...
Solution:
Given, the tension in the sonometer wire is increased by 69%, keeping the length constant.
Let the initial tension in the wire be T and the frequency of the wire be f.
After increasing the tension by 69%, the new tension in the wire is given by:
T' = T + (69/100)T = (169/100)T
Now, using the formula for the fundamental frequency of a stretched string, we have:
f' = (1/2L) * sqrt(T'/m)
Since the length of the wire is constant, we can write:
f' = k * sqrt(T')
where k is a constant.
Substituting T' = (169/100)T, we get:
f' = k * sqrt((169/100)T)
Also, we know that the frequency of the wire increases by 9 Hz.
Therefore, we can write:
f' - f = 9
Substituting the expression for f' from above, we get:
k * sqrt((169/100)T) - k * sqrt(T) = 9
Taking k as common, we get:
k * (sqrt((169/100)T) - sqrt(T)) = 9
Dividing both sides by k and squaring, we get:
((169/100)T - T) = (81/k^2)
Simplifying, we get:
T = (81/k^2) * (100/69)
Substituting this value of T in the expression for f', we get:
f' = k * sqrt((169/100) * (81/k^2) * (100/69))
Simplifying, we get:
f' = (3/2) * k
Since we know that f' - f = 9, we can write:
(3/2) * k - k = 9
Solving for k, we get:
k = 18
Substituting this value of k in the expression for f', we get:
f' = (3/2) * k = 27
Therefore, the frequency of the wire is 27 Hz.
Hence, the correct answer is option (c) 30 Hz.
Given, the tension in the sonometer wire is increased by 69%, keeping the length constant.
Let the initial tension in the wire be T and the frequency of the wire be f.
After increasing the tension by 69%, the new tension in the wire is given by:
T' = T + (69/100)T = (169/100)T
Now, using the formula for the fundamental frequency of a stretched string, we have:
f' = (1/2L) * sqrt(T'/m)
Since the length of the wire is constant, we can write:
f' = k * sqrt(T')
where k is a constant.
Substituting T' = (169/100)T, we get:
f' = k * sqrt((169/100)T)
Also, we know that the frequency of the wire increases by 9 Hz.
Therefore, we can write:
f' - f = 9
Substituting the expression for f' from above, we get:
k * sqrt((169/100)T) - k * sqrt(T) = 9
Taking k as common, we get:
k * (sqrt((169/100)T) - sqrt(T)) = 9
Dividing both sides by k and squaring, we get:
((169/100)T - T) = (81/k^2)
Simplifying, we get:
T = (81/k^2) * (100/69)
Substituting this value of T in the expression for f', we get:
f' = k * sqrt((169/100) * (81/k^2) * (100/69))
Simplifying, we get:
f' = (3/2) * k
Since we know that f' - f = 9, we can write:
(3/2) * k - k = 9
Solving for k, we get:
k = 18
Substituting this value of k in the expression for f', we get:
f' = (3/2) * k = 27
Therefore, the frequency of the wire is 27 Hz.
Hence, the correct answer is option (c) 30 Hz.
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The fundamental frequency of the sonometer wire Increases by 9 Hz, if its tension is increased by 69%, keeping the length constant. The frequency of the wire isa)42 Hzb)24 Hzc)30 Hzd)36 HzCorrect answer is option 'C'. Can you explain this answer?
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