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Waves
Beats
9 questions from 9 papers, each with its answer.
- questions
- 9
- papers
- 9
- years asked, 2017–2026
- 1 of 10
- of Waves
- 15.25%
Appears in
Concepts tested
- Beat frequency 3
- Beat frequency and intensity ratio
- beat frequency and tension
- Beat frequency from wave equations
- Frequency of a stretched string
- Fundamental frequency of a string
- three-wave-beats
All Questions
2020 1 question
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In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6 Hz. When tension in B is slightly decreased, the beat frequency increases to 7 Hz. If the frequency of A is 530 Hz, the original frequency of B will be:Initially $|f_A - f_B| = 6$ Hz. When tension in B is decreased, $f_B$ decreases. Since the beat frequency increases to 7 Hz, $f_A > f_B$. $f_A - f_B = 6 \Rightarrow f_B = 530 - 6 = 524$ Hz (original)
2016 Phase II 1 question
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Three sound waves of equal amplitudes have frequencies (n – 1), n, (n + 1). They superimpose to give beats. The number of beats produced per second will be:
The resultant repeats with the largest frequency difference, 2 Hz.
Net beat frequency = LCM of individual beat frequencies = LCM of [(n, n–1), (n, n+1), (n–1, n+1)] = LCM of (1, 1, 2) = 2 Hz
So the number of beats per second = 2
2013 NEET 1 question
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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 isFrequency of the unknown source = 246 Hz or 254 Hz
Its second harmonic = 492 Hz or 508 Hz
Only 508 Hz gives 5 beats per second with 513 Hz, so the unknown frequency = 254 Hz
2012 AIPMT-PRE 1 question
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Two sources of sound placed close to each other, are emitting progressive waves given by $y_1 = 4\sin600\pi t$ and $y_2 = 5\sin608\pi t$. An observer located near these two sources of sound will hear:$\omega_1 = 600\pi \Rightarrow f_1 = 300$ Hz; $\omega_2 = 608\pi \Rightarrow f_2 = 304$ Hz
Beat frequency $= 304 - 300 = 4$ beats per second
$\frac{I_{max}}{I_{min}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = \frac{(4 + 5)^2}{(5 - 4)^2} = \frac{81}{1}$
So the observer hears 4 beats per second with intensity ratio 81 : 1.
2011 AIPMT-MAINS 1 question
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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 beFundamental frequency of a stretched wire: $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, so $f \propto \sqrt{T}$
For a small change: $\frac{\Delta f}{f} = \frac{1}{2}\frac{\Delta T}{T}$
6 beats per second means the frequencies differ by $\Delta f = 6$ Hz.
$\frac{6}{600} = \frac{1}{2}\frac{\Delta T}{T}$
$\frac{\Delta T}{T} = 2\times\frac{6}{600} = \frac{1}{50}$
$\frac{\Delta T}{T} = 0.02$
2010 AIPMT-PRE 1 question
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A tuning fork of frequency 512 Hz makes 4 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per sec when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension wasBeat frequency = difference of the two frequencies, so the string frequency is either 512 + 4 = 516 Hz or 512 − 4 = 508 Hz.
Frequency of a string $f \propto \sqrt{T}$, so increasing the tension increases its frequency.
If the string were at 516 Hz, raising its frequency would increase the difference from 512 Hz, and the beats would increase.
If the string were at 508 Hz, raising its frequency (to 510 Hz) brings it closer to 512 Hz, and the beats decrease to 2 per second, as observed.
Hence the original frequency of the piano string was 508 Hz.
2009 AIPMT 1 question
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Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously the number of beats isFundamental frequency of a stretched string: $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$
T = 20 N, $\mu$ = 1 g/m = $10^{-3}$ kg/m: $\sqrt{\frac{T}{\mu}} = \sqrt{\frac{20}{10^{-3}}} = \sqrt{2\times10^4} = 141.4$ m/s
$f_1 = \frac{141.4}{2\times0.516} = 137.0$ Hz
$f_2 = \frac{141.4}{2\times0.491} = 144.0$ Hz
Number of beats per second = $f_2 - f_1 = 144 - 137$
= 7
2006 AIPMT-PRE 1 question
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Two sound waves with wavelengths 5.0 m and 5.5 m respectively, each propagate in a gas with velocity 330 m/s. We expect the following number of beats per second :-Frequency of a wave: $n = \dfrac{v}{\lambda}$
$n_1 = \dfrac{330}{5.0} = 66$ Hz
$n_2 = \dfrac{330}{5.5} = 60$ Hz
Number of beats per second $= n_1 - n_2 = \dfrac{v}{\lambda_1} - \dfrac{v}{\lambda_2} = 330\left(\dfrac{1}{5} - \dfrac{1}{5.5}\right)$
$= 66 - 60$
$= 6$
2005 AIPMT 1 question
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Two vibrating tuning forks produce progressive waves given by $Y_1 = 4\sin 500\pi t$ and $Y_2 = 2\sin 506\pi t$. Number of beats produced per minute isCompare each equation with $Y = A\sin 2\pi nt$.
$2\pi n_1 = 500\pi \Rightarrow n_1 = 250$ Hz
$2\pi n_2 = 506\pi \Rightarrow n_2 = 253$ Hz
Number of beats per second $= n_2 - n_1 = 253 - 250 = 3$
Number of beats produced per minute $= 3 \times 60 = 180$