Beats
When two sound waves of slightly different frequencies, travelling in a medium along the
same direction, superimpose on each other, the intensity of the resultant sound at a particular
position rises and falls regularly with time. This phenomenon of regular variation in intensity of
sound with time at a particular position is called beats.
(1) Persistence of hearing : The impression of sound heard by our ears persist in our mind
for 1/10th of a second. If another sound is heard before 1/10 second is over, the impression of the
two sound mix up and our mind cannot distinguish between the two.
So for the formation of distinct beats, frequencies of two sources of sound should be nearly
equal (difference of frequencies less than 10)
(2) Equation of beats : If two waves of equal amplitudes 'a' and slightly different
frequencies n1 and n2 travelling in a medium in the same direction are.
ω 1 t = a sin 2 πn1 t y 2 = a sin ω2 t=asin 2 π n 2 t
y1 = a sin ;
⃗
y =⃗
y 1 +⃗
y2
By the principle of super position :
π (n +n ) t π (n −n ) t
y = A sin 1 2
where A = 2a cos 1 2
= Amplitude of resultant wave.
(3) One beat : If the intensity of sound is maximum at time t = 0, one beat is said to be
formed when intensity becomes maximum again after becoming minimum once in between.
(4) Beat period : The time interval between two successive beats (i.e. two successive
maxima of sound) is called beat period.
n=n1 ~ n2
(5) Beat frequency : The number of beats produced per second is called beat frequency.
1 1
T= =
Beat frequency n1 ~ n2
Determination of Unknown Frequency
Suppose a tuning fork of known frequency (nA) is sounded together with another tuning fork
of unknown frequency (nB) and x beats heard per second.
A B
Kno Unkn
x
Fig.
There are two possibilities to known frequency of unknown tuning fork.
n A −n B =x
... (i)
n B −n A =x
or ... (ii)
To find the frequency of unknown tuning fork (nB) following steps are taken.
(1) Loading or filing of one prong of known or unknown (by wax) tuning fork, so frequency
changes (decreases after loading, increases after filing).
(2) Sound them together again, and count the number of heard beats per sec again, let it be x.
These are following four condition arises.
(i) x > x (ii) x < x (iii) x = 0 (iv) x = x
(3) With the above information, the exact frequency of the unknown tuning fork can be
determined as illustrated below.
Suppose two tuning forks A (frequency nA is known) and B (frequency nB is unknown) are
sounded together and gives x beats/sec. If one prong of unknown tuning fork B is loaded with a
little wax (so nB decreases) and it is sounded again together with known tuning fork A, then in the
following four given condition nB can be determined.
(4) If x > x than x, then this would happen only when the new frequency of B is more away
from nA. This would happen if originally (before loading), nB was less than nA.
Thus initially nB = nA – x.
(5) If x < x than x, then this would happen only when the new frequency of B is more nearer
to nA. This would happen if originally (before loading), nB was more than nA.
Thus initially nB = nA + x.
(6) If x = x then this would means that the new frequency (after loading) differs from nA by
the same amount as was the old frequency (before loading). This means initially nB = nA + x
(and now it has decreased to nB = nA – x)
(7) If x = 0, then this would happen only when the new frequency of B becomes equal to nA.
This would happen if originally nB was more than nA.
Thus initially nB = nA + x.
Table 17.7 ; Frequency of unknown tuning fork for various cases
By loading
If B is loaded with wax so its frequency If A is loaded with wax its frequency
decreases decreases
If x increases nB = nA – x If x increases nB = nA + x
If x decrease nB = nA + x If x decrease nB = nA – x
If remains same nB = nA + x If remains same nB = nA – x
If x becomes zero nB = nA + x If x becomes zero nB = nA – x
By filing
If B is filed, its frequency increases If A is filed, its frequency increases
If x increases nB = nA + x If x increases nB = nA – x
If x decrease nB = nA – x If x decrease nB = nA + x
If remains same nB = nA – x If remains same nB = nA + x
If x becomes zero nB = nA – x If x becomes zero nB = nA + x
