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Ten tuning fork are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce 4 beats per second. The highest frequency is twice that of the lowest. Possible highest and lowest frequencies are (a) 80 & 40 (b) 100 & 50 (c) 44 & 22 (d) 72 & 36

Ten tuning fork are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce 4 beats per second. The highest frequency is twice that of the lowest. Possible highest and lowest frequencies are
(a) 80 & 40
(b) 100 & 50
(c) 44 & 22
(d) 72 & 36

Grade:12

1 Answers

Vikas TU
14149 Points
one year ago
Let the frequency of first tuning fork is x then then we add on 4 to x to get frequency of next fork so by doing this the frequency of the last fork will be 36+x .
Now since it is given that the frequency of last tuning fork is double than that of first we get other frequency of last tuning fork = 2x
Now by solving these two equations we get
x = 36
therefore least frequency is 36 and max. is 72.
 

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