Beats

1. Two tuning forks when sounded together produced 4 beats/sec. The frequency of one fork
is 256. The number of beats heard increases when the fork of frequency 256 is loaded with
wax. The frequency of the other fork is
(a) 504 (b) 520
(c) 260 (d) 252
2. Beats are the result of
(a) Diffraction
(b) Destructive interference
(c) Constructive and destructive interference
(d) Superposition of two waves of nearly equal frequency
3. Two adjacent piano keys are struck simultaneously. The notes emitted by them have
n1 n2
frequencies and . The number of beats heard per second is
1 1
(n 1−n 2 ) (n 1 +n2 )
(a) 2 (b) 2
n1 ~n 2 2 (n 1−n 2 )
(c) (d)
4. A tuning fork of frequency 100 when sounded together with another tuning fork of
unknown frequency produces 2 beats per second. On loading the tuning fork whose
frequency is not known and sounded together with a tuning fork of frequency 100 produces
one beat, then the frequency of the other tuning fork is
(a) 102 (b) 98
(c) 99 (d) 101
5. A tuning fork sounded together with a tuning fork of frequency 256 emits two beats. On
loading the tuning fork of frequency 256, the number of beats heard are 1 per second. The
frequency of tuning fork is
(a) 257 (b) 258
(c) 256 (d) 254
6. If two tuning forks A and B are sounded together, they produce 4 beats per second. A is then
slightly loaded with wax, they produce 2 beats when sounded again. The frequency of A is
256. The frequency of B will be
(a) 250 (b) 252
(c) 260 (d) 262
7. The frequencies of two sound sources are 256 Hz and 260 Hz. At t = 0, the intensity of
sound is maximum. Then the phase difference at the time t = 1/16 sec will be
(a) Zero (b) 
(c) /2 (d) /4
8. Two tuning forks have frequencies 450 Hz and 454 Hz respectively. On sounding these
forks together, the time interval between successive maximum intensities will be
(a) 1/4 sec (b) 1/2 sec
(c) 1 sec (d) 2 sec
9. When a tuning fork of frequency 341 is sounded with another tuning fork, six beats per
second are heard. When the second tuning fork is loaded with wax and sounded with the
first tuning fork, the number of beats is two per second. The natural frequency of the second
tuning fork is
(a) 334 (b) 339
(c) 343 (d) 347
10. Two tuning forks of frequencies 256 and 258 vibrations/sec are sounded together, then time
interval between consecutive maxima heard by the observer is
(a) 2 sec (b) 0.5 sec
(c) 250 sec (d) 252 sec

11. A tuning fork gives 5 beats with another tuning fork of frequency 100 Hz. When the first
tuning fork is loaded with wax, then the number of beats remains unchanged, then what will
be the frequency of the first tuning fork
(a) 95 Hz (b) 100 Hz
(c) 105 Hz (d) 110 Hz
F1
12. Tuning fork has a frequency of 256 Hz and it is observed to produce 6 beats/second with
F2 F2 F1
another tuning fork . When is loaded with wax, it still produces 6 beats/second with
F
. The frequency of 2 before loading was
(a) 253 Hz (b) 262 Hz
(c) 250 Hz (d) 259 Hz
13. A tuning fork and a sonometer wire were sounded together and produce 4 beats per second.
When the length of sonometer wire is 95 cm or 100 cm, the frequency of the tuning fork is

(a) 156 Hz (b) 152 Hz

(c) 148 Hz (d) 160 Hz
14. Two tuning forks A and B vibrating simultaneously produce 5 beats. Frequency of B is 512.
It is seen that if one arm of A is filed, then the number of beats increases. Frequency of A
will be
(a) 502 (b) 507
(c) 517 (d) 522
15. The beats are produced by two sound sources of same amplitude and of nearly equal
frequencies. The maximum intensity of beats will be ...... that of one source
(a) Same (b) Double
(c) Four times (d) Eight times
 y =a sin 2000 πt y =a sin 2008 πt
16. Beats are produced by two waves given by 1 and 2 . The number
of beats heard per second is
(a) Zero (b) One
(c) Four (d) Eight
17. A tuning fork whose frequency as given by manufacturer is 512 Hz is being tested with an
accurate oscillator. It is found that the fork produces a beat of 2 Hz when oscillator reads
514 Hz but produces a beat of 6 Hz when oscillator reads 510 Hz. The actual frequency of
fork is
(a) 508 Hz (b) 512 Hz
(c) 516 Hz (d) 518 Hz
18. A tuning fork of frequency 480 Hz produces 10 beats per second when sounded with a
vibrating sonometer string. What must have been the frequency of the string if a slight
increase in tension produces lesser beats per second than before
(a) 460 Hz (b) 470 Hz
(c) 480 Hz (d) 490 Hz
19. When a tuning fork A of unknown frequency is sounded with another tuning fork B of
frequency 256 Hz, then 3 beats per second are observed. After that A is loaded with wax
and sounded, the again 3 beats per second are observed. The frequency of the tuning fork A
is
(a) 250 Hz (b) 253 Hz
(c) 259 Hz (d) 262 Hz
20. A source of sound gives five beats per second when sounded with another source of
−1
frequency 100 s . The second harmonic of the source together with a source of frequency
205 s−1 gives five beats per second. What is the frequency of the source
−1 −1
(a) 105 s (b) 205 s
−1 −1
(c) 95 s (d) 100 s
21. When two sound waves are superimposed, beats are produced when they have

(a) Different amplitudes and phases

(b) Different velocities
(c) Different phases
(d) Different frequencies
22. Two tuning forks A and B give 4 beats per second. The frequency of A is 256 Hz. On
loading B slightly, we get 5 beats in 2 seconds. The frequency of B after loading is
(a) 253.5 Hz (b) 258.5 Hz
(c) 260 Hz (d) 252 Hz
23. A tuning fork A of frequency 200 Hz is sounded with fork B, the number of beats per
second is 5. By putting some wax on A, the number of beats increases to 8. The frequency
of fork B is
 (a) 200 Hz (b) 195 Hz
(c) 192 Hz (d) 205 Hz
24. Two tuning forks, A and B, give 4 beats per second when sounded together. The frequency
of A is 320 Hz. When some wax is added to B and it is sounded with A, 4 beats per second
are again heard. The frequency of B is
(a) 312 Hz (b) 316 Hz
(c) 324 Hz (d) 328 Hz
25. Two tuning forks have frequencies 380 and 384 Hz respectively. When they are sounded
together, they produce 4 beats. After hearing the maximum sound, how long will it take to
hear the minimum sound
1 1
(a) 2 sec (b) 4 sec
1 1
(c) 8 sec (d) 16 sec
26. Beats are produced with the help of two sound waves of amplitudes 3 and 5 units. The ratio
of maximum to minimum intensity in the beats is
(a) 2 : 1 (b) 5 : 3
(c) 4 : 1 (d) 16 : 1
27. Two waves of lengths 50 cm and 51 cm produced 12 beats per second. The velocity of
sound is
(a) 306 m/s (b) 331 m/s
(c) 340 m/s (d) 360 m/s
28. Two waves y=0.25sin 316 t and y=0.25sin 310 t are travelling in same direction. The number
of beats produced per second will be
(a) 6 (b) 3
(c) 3/ (d) 3
29. The couple of tuning forks produces 2 beats in the time interval of 0.4 seconds. So the beat
frequency is
(a) 8 Hz (b) 5 Hz
(c) 2 Hz (d) 10 Hz
30. An unknown frequency x produces 8 beats per seconds with a frequency of 250 Hz and 12
beats with 270 Hz source, then x is
(a) 258 Hz (b) 242 Hz
(c) 262 Hz (d) 282 Hz
 Beats

1 c 2 d 3 c 4 a 5 d
6 b 7 c 8 a 9 d 10 b
11 c 12 b 13 a 14 c 15 c
16 c 17 c 18 b 19 c 20 a
21 d 22 c 23 d 24 c 25 c
26 d 27 a 28 c 29 b 30 a

Beats

n A =256 Hz
1. (c) Suppose two tuning forks are named A and B with frequencies (known), nB =
? (unknown), and beat frequency x = 4 bps.
Known
Unknown
nA nB
A A
x bps

Frequency of unknown tuning fork may be

n B =256+ 4=260 Hz

or =256−4=252 Hz
It is given that on sounding waxed fork A (fork of frequency 256 Hz) and fork B,
number of beats (beat frequency) increases. It means that with decrease in frequency of
A, the difference in new frequency of A and the frequency of B has increased. This is
possible only when the frequency of A while decreasing is moving away from the
frequency of B.
This is possible only if nB = 260 Hz.
 n =256 Hz , n =?
Alternate method : It is given A B
and x = 4 bps
Also after loading A (i.e. nA ), beat frequency (i.e. x) increases ().
Apply these informations in two possibilities to known the frequency of unknown
tuning fork.
nA  – nB = x ... (i)
nB – nA  = x ... (ii)
It is obvious that equation (i) is wrong (ii) is correct so
nB = nA + x = 256 + 4 = 260 Hz.
2. (d)
3. (c)
4. (a) Suppose nA = known frequency = 100 Hz, nB = ?
x = 2 = Beat frequency, which is decreasing after loading (i.e. x)
Unknown tuning fork is loaded so nB
Hence nA – nB  = x ... (i) Wrong
nB  – nA = x ... (ii) Correct
 nB = nA + x = 100 + 2 = 102 Hz.
5. (d) nA = Known frequency = 256, nB = ?
x = 2 bps, which is decreasing after loading (i.e. x) known tuning fork is loaded so nA
Hence nA – nB = x ... (i) Correct
nB – nA = x ... (ii) Wrong
 nB = nA – x = 256 – 2 = 254 Hz.
6. (b) nA = Known frequency = 256 Hz, nB = ?
x = 4 bps, which is decreasing after loading (i.e. x) also known tuning fork is loaded so
nA
Hence nA – nB = x ... (i) Correct
nB – nA = x ... (ii) Wrong
 nB = nA – x = 256 – 4 = 252 Hz.
7. (c) Time interval between two consecutive beats
1 1 1 1 T
T= = = sec t= = sec
n1 −n2 260−256 4 16 4
so,
T
׿ ¿
By using time difference = 2π Phase difference
T T π
= ×φ⇒ φ=
⇒4 2π 2
8. (a) The time interval between successive maximum intensities will be
1 1 1
= = sec .
n1 ~ n2 454−450 4
9. (d) nA = Known frequency = 341 Hz, nB = ?
x = 6 bps, which is decreasing (i.e. x) after loading (from 6 to 1 bps)
Unknown tuning fork is loaded so nB
Hence nA – nB = x ... (i) Wrong
nB – nA = x ... (ii) Correct
  nB = nA + x = 341 + 6 = 347 Hz.
1
T=
10. (b) 258−256 = 0 . 5 sec
11. (c) Suppose nA = known frequency = 100 Hz, nB = ?
x = 5 bps, which remains unchanged after loading
Unknown tuning fork is loaded so nB
Hence nA – nB  = x ... (i)
nB  – nA = x ... (ii)
From equation (i), it is clear that as nB decreases, beat frequency. (i.e. nA – (nB)new) can
never be x again.
From equation (ii), as nB, beat frequency (i.e. (nB)new – nA) decreases as long as (nB)new
remains greater than nA, If (nB)new become lesser than nA the beat frequency will increase
again and will be x. Hence this is correct.
So, nB = nA + x = 100 + 5 = 105 Hz.
12. (b) nA = Known frequency = 256 Hz, nB = ?
x = 6 bps, which remains the same after loading.
Unknown tuning fork F2 is loaded so nB
Hence nA – nB = x ... (i) Wrong
nB – nA = x ... (ii) Correct
 nB = nA + x = 256 + 6 = 262 Hz.
13. (a) Probable frequencies of tuning fork be n+ 4 or n−4
1
n∝
Frequency of sonometer wire l
n+ 4 100
=
∴ n−4 95 or 95 (n+4)=100(n−4)

or 95 n+380=100 n−400 or 5 n=780 or n=156

n
14. (c) After filling frequency increases, so A decreases (). Also it is given that beat
frequency increases (i.e., x )
Hence nA – nB = x ... (i) Correct
nB – nA = x ... (ii) Wrong
 nA = nB + x = 512 + 5 = 517 Hz.
15. (c) Intensity  (amplitude)2
A max =2 ao I max =4 I o
as (ao= amplitude of one source) so .
n1 ~n 2
16. (c) Number of beats per second =
ω 1=2000 π=2 πn1
 n1 = 1000
ω 2=2008 π =2 πn2
and  n2 = 1004
Number of beats heard per sec=1004−1000=4
17. (c) The tuning fork whose frequency is being tested produces 2 beats with oscillator at 514
Hz, therefore, frequency of tuning fork may either be 512 or 516. With oscillator
frequency 510 it gives 6 beats/sec, therefore frequency of tuning fork may be either 516
or 504.
Therefore, the actual frequency is 516 Hz which gives 2 beats/sec with 514 Hz and 6
beats/sec with 510 Hz.

18. (b) If suppose nS = frequency of string

=
√
1 T
2l m
nf = Frequency of tuning fork = 480 Hz
x = Beats heard per second = 10
as tension T increases, so nS increases ()
Also it is given that number of beats per sec decreases (i.e. x)
Hence nS – nf = x ... (i) Wrong
nf – nS = x ... (ii) Correct
 nS = nf – x = 480 – 10 = 470 Hz.
19. (c) It is given that
nA = Unknown frequency = ?
nB = Known frequency = 256 Hz
x = 3 bps, which remains same after loading
Unknown tuning fork A is loaded so nA
Hence nA – nB = x ... (i) Correct
nB – nA = x ... (ii) Wrong
 nA = nB + x = 256 + 3 = 259 Hz.
20. (a) Frequency of the source = 100  5 = 105 Hz or 95 Hz.
Second harmonic of the source = 210 Hz or 190 Hz.
As the second harmonic gives 5 beats/sec with sound of frequency 205 Hz, the second
harmonic should be 210 Hz.
 Frequency of the source = 105 Hz.
21. (d) For producing beats, their must be small difference in frequency.
22. (c) nA = Known frequency = 256 Hz, nB = ?
5
bps
x = 4 beats per sec which is decreasing (4 bps to 2 ) after loading (i.e. x)
Unknown tuning fork B, is loaded so nB
Hence nA – nB = x ... (i) Wrong
nB – nA = x ... (ii) Correct
 nB = nA + x = 256 + 4 = 260 Hz.
23. (d) nA – nB = x ... (i) Wrong
nB – nA = x ... (ii) Correct
 nB = nA + x = 200 + 5 = 205 Hz.
24. (c) nA – nB = x (same) ... (i) Wrong
nB – nA = x (same) ... (ii) Correct
  nB = nA + x = 320 + 4 = 324 Hz.
1 1 1
T= = = sec
n1 ~ n2 384−380 4
25. (c) Beat period . Hence minimum time interval between maxima
T 1
t= = sec .
and minima 2 8

( )
I max a1 + a2 2 (5+3 )2 16
= = =
I min a1 −a2 (5−3 )2 1
26. (d)
v v v v
n1 = = n2 = =
λ1 0 . 50 λ2 0 . 51
27. (a) and
Δn=n 1−n 2=v
1
−
[ 1
0. 05 0 . 51
=12
]
12×0 . 51×0 . 50
v= =306
 0 .01 m/s
316 310 316 310 3
n1 = f 2= − =
28. (c) 2 π and 2 π Number of beats heard per second = n – n = 2 π 2 π π
1 2
2
=5 Hz
29. (b) Beat frequency = 0 . 4
30. (a) Since source of frequency x gives 8 beats per second with frequency 250 Hz, it's
possible frequency are 258 or 242. As source of frequency x gives 12 beats per second
with a frequency 270 Hz, it's possible frequencies 282 or 258 Hz. The only possible
frequency of x which gives 8 beats with frequency 250 Hz also 12 beats per second with
258 Hz.