Waves
- A string is stretched between two fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is :
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In a stretched string all multiples of frequencies can be obtained i.e., if fundamental frequency is n then higher frequencies will be 2n, 3n, 4n ...
So, the difference between any two successive frequencies will be 'n'
According to question, n = 420 – 315 = 105 Hz
So the lowest frequency of the string is105 Hz.Correct Option: C
In a stretched string all multiples of frequencies can be obtained i.e., if fundamental frequency is n then higher frequencies will be 2n, 3n, 4n ...
So, the difference between any two successive frequencies will be 'n'
According to question, n = 420 – 315 = 105 Hz
So the lowest frequency of the string is105 Hz.
- If n1, n2 and n3 are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by :
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n = 1 √ T 2l m or, n ∝ 1 or nl = constant, K l
∴ n1l1 = K,
n2l2 = K, n3l3 = K
Also, l = l1 + l2 +l3or, K = K + K + K n n1 n2 n3 or, 1 = 1 + 1 + 1 n n1 n2 n3 Correct Option: A
n = 1 √ T 2l m or, n ∝ 1 or nl = constant, K l
∴ n1l1 = K,
n2l2 = K, n3l3 = K
Also, l = l1 + l2 +l3or, K = K + K + K n n1 n2 n3 or, 1 = 1 + 1 + 1 n n1 n2 n3
- The number of possible natural oscillation of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are : (velocity of sound = 340 ms-1)
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In case of closed organ pipe frequency,
fn = (2n + 1)
for n = 0, f0 = 100 Hz
n = 1, f1 = 300 Hz
n = 2, f2 = 500 Hz
n = 3, f3 = 700 Hz
n = 4, f4 = 900 Hz
n = 5, f5 = 1100 Hz
n = 6, f6 = 1300 Hz
Hence possible natural oscillation whose frequencies < 1250 Hz = 6(n = 0, 1, 2, 3, 4, 5)Correct Option: D
In case of closed organ pipe frequency,
fn = (2n + 1)
for n = 0, f0 = 100 Hz
n = 1, f1 = 300 Hz
n = 2, f2 = 500 Hz
n = 3, f3 = 700 Hz
n = 4, f4 = 900 Hz
n = 5, f5 = 1100 Hz
n = 6, f6 = 1300 Hz
Hence possible natural oscillation whose frequencies < 1250 Hz = 6(n = 0, 1, 2, 3, 4, 5)
- If we study the vibration of a pipe open at both ends, then which of the following statements is not true ?
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Pressure change will be minimum at both ends. In fact, pressure variation is maximum at l/2 because the displacement node is pressure antinode.
Correct Option: C
Pressure change will be minimum at both ends. In fact, pressure variation is maximum at l/2 because the displacement node is pressure antinode.
- The length of the wire between two ends of a sonometer is 100 cm. What should be the positions of two bridges below the wire so that the three segments of the wire have their fundamental frequencies in the ratio of 1 : 3 : 5?
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From formula, ƒ = 1 √ T x m ⇒ 1 ∝ l ƒ ∴ l1 : l2 : l3 = 1 : 1 : 1 ƒ1 ƒ2 ƒ3
= f2 f3 : f1 f3 : f1 f2
[Given: f1 : f2 : f3 = 1 : 3 : 5]
= 15 : 5 : 3
Therefore the positions of two bridges below the wire are15 × 100 cm and 15 × 100 + 15 × 100 cm 15 + 5 + 3 15 + 5 + 3 i.e., 1500 cm, 2000 cm 23 23 Correct Option: A
From formula, ƒ = 1 √ T x m ⇒ 1 ∝ l ƒ ∴ l1 : l2 : l3 = 1 : 1 : 1 ƒ1 ƒ2 ƒ3
= f2 f3 : f1 f3 : f1 f2
[Given: f1 : f2 : f3 = 1 : 3 : 5]
= 15 : 5 : 3
Therefore the positions of two bridges below the wire are15 × 100 cm and 15 × 100 + 15 × 100 cm 15 + 5 + 3 15 + 5 + 3 i.e., 1500 cm, 2000 cm 23 23
