Waves Difficult Questions and Answers | Page - 14

Waves

A closed organ pipe (closed at one end) is excited to support the third overtone. It is found that air in the pipe has

1. three nodes and three antinodes
1. three nodes and four antinodes
1. four nodes and three antinodes
1. four nodes and four antinodes

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Third overtone has a frequency 7 n, which means L = 7λ/4 = three full loops + one half loop, which would make four nodes and four antinodes.

Correct Option: D

Third overtone has a frequency 7 n, which means L = 7λ/4 = three full loops + one half loop, which would make four nodes and four antinodes.

A source of unknown frequency gives 4 beats/s, when sounded with a source of known frequency 250 Hz. The second harmonic of the source of unknown frequency gives five beats per second, when sounded with a source of frequency 513 Hz. The unknown frequency is

1. 246 Hz
1. 240 Hz
1. 260 Hz
1. 254 Hz

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When sounded with a source of known frequency fundamental frequency
= 250 ± 4 Hz = 254 Hz or 246 Hz
2^nd harmonic if unknown frequency (suppose) 254
Hz = 2 × 254 = 508 Hz
As it gives 5 beats
∴ 508 + 5 = 513 Hz
Hence, unknown frequency is 254 Hz

Correct Option: D

When sounded with a source of known frequency fundamental frequency
= 250 ± 4 Hz = 254 Hz or 246 Hz
2^nd harmonic if unknown frequency (suppose) 254
Hz = 2 × 254 = 508 Hz
As it gives 5 beats
∴ 508 + 5 = 513 Hz
Hence, unknown frequency is 254 Hz

Two sources P and Q produce notes of frequency 660 Hz each. A listener moves from P to Q with a speed of 1 ms^–1. If the speed of sound is 330 m/s, then the number of beats heard by the listener per second will be

1. zero
1. 4
1. 8
1. 2

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Δƒ - v
ƒ C

⇒ (Beats)/2 = v
ƒ C

⇒ Beats = 2ƒv = 4
C

Correct Option: B

Δƒ - v
ƒ C

⇒ (Beats)/2 = v
ƒ C

⇒ Beats = 2ƒv = 4
C

Two sources of sound placed close to each other are emitting progressive waves given by y₁ = 4 sin 600 πt and y₂ = 5 sin 608 πt. An observer located near these two sources of sound will hear :

1. 4 beats per second with intensity ratio 25 : 16 between waxing and waning.
1. 8 beats per second with intensity ratio 25 : 16 between waxing and waning
1. 8 beats per second with intensity ratio 81 : 1 between waxing and waning
1. 4 beats per second with intensity ratio 81 : 1 between waxing and waning

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2π ƒ₁ = 600 π
ƒ₁ = 300 ... (1)
2π ƒ₂ = 608 π
ƒ₂ = 304 ... (2)
|ƒ₁ – ƒ₂| = 4 beats
I_max = (A₁ + A₂)² = (5 + 4)² = 81
I_min (A₁ + A₁)² (5 - 4)² 1

where A₁,A₁ are amplitudes of given two sound wave.

Correct Option: D

2π ƒ₁ = 600 π
ƒ₁ = 300 ... (1)
2π ƒ₂ = 608 π
ƒ₂ = 304 ... (2)
|ƒ₁ – ƒ₂| = 4 beats
I_max = (A₁ + A₂)² = (5 + 4)² = 81
I_min (A₁ + A₁)² (5 - 4)² 1

where A₁,A₁ are amplitudes of given two sound wave.

Two identical piano wires kept under the same tension T have a fundamental frequency of 600 Hz. The fractional increase in the tension of one of the wires which will lead to occurrence of 6 beats/s when both the wires oscillate together would be

1. 0.02
1. 0.03
1. 0.04
1. 0.01

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For fundamental mode,
f = 1 √ T
2l m

Taking logarithm on both sides, we get
log f = log 1 + log √ T
2l μ

= log 1 + 1 log T
2l 2 μ

or log f = log l + 1 [logT - logμ]
2l 2

Differentiating both sides, we get
df = 1 dT (as and µ are constants)
f 2 T

⇒ dT = 2 × df
T T

Here df = 6
f = 600 Hz
∴ dT = 2 × 6 = 0.02
T 600

Correct Option: A

For fundamental mode,
f = 1 √ T
2l m

Taking logarithm on both sides, we get
log f = log 1 + log √ T
2l μ

= log 1 + 1 log T
2l 2 μ

or log f = log l + 1 [logT - logμ]
2l 2

Differentiating both sides, we get
df = 1 dT (as and µ are constants)
f 2 T

⇒ dT = 2 × df
T T

Here df = 6
f = 600 Hz
∴ dT = 2 × 6 = 0.02
T 600