Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
Frequency of string A, fA = 324 Hz
Frequency of string B = fB
Beat’s frequency, n = 6 Hz
beat's frequency is given as :
\(n=|f_A±f_B|\)
\(6=324±f_B\)
\(f_B=330\, Hz \,or \,318\, Hz\)
Frequency decreases with a decrease in the tension in a string. This is because frequency is directly proportional to the square root of tension. It is given as:
\(v∝√T\)
Hence the beat frequency cannot be 330 Hz
\(∴f_B=318 Hz\)
Give reasons for the following.
(i) King Tut’s body has been subjected to repeated scrutiny.
(ii) Howard Carter’s investigation was resented.
(iii) Carter had to chisel away the solidified resins to raise the king’s remains.
(iv) Tut’s body was buried along with gilded treasures.
(v) The boy king changed his name from Tutankhaten to Tutankhamun.
Find the mean deviation about the median for the data
xi | 15 | 21 | 27 | 30 | 35 |
fi | 3 | 5 | 6 | 7 | 8 |
Let’s see from fig.m, the frequency of a pink-colored wave is f1, and that of a green-colored wave is f2. So, the frequency of the beat is the difference between these two, which is:
fBEATS = |f1 - f2|
When two or more waves travelling in a medium meet, the resulting phenomenon is called interference and beats are an excellent example of the phenomenon of interference.
When two or more waves travelling in a medium meet, the resulting phenomenon is called interference and beats are an excellent example of the phenomenon of interference.