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Question
A set of 48 tuning forks is arranged in a series of descending frequencies such that each fork gives 4 beats per second with preceding one. The frequency of first fork is 1.5 times the frequency of the last fork, find the frequency of the first and 42nd tuning fork
Solution
Given that :
n1=1.5 n48 ,
beat frequency =4Hz
The set of tuning forks are arranged in decreasing order of frequencies.
∴n2=n1-4
n3=n2-4=n1-2 × 4
n48=n47-4=n1-47 × 4
∴n48=n1-188
∴n48=1.5n48-188 ........(n1=1.5n48)
∴0.5n48=188
∴n48=376
→n1=1.5n48=1.5 × 376=564
n42=n41-4=n1
Given that n1=1.5 n48 and beat frequency =5Hz
The set of tuning forks are arranged in decreasing order of frequencies.
∴n2=n1-4
n3=n2-4=n1-2 × 4
n48=n47-4=n1-47 × 4
∴n48=n1-188
∴n48=1.5n48-188 ........(n1=1.5n48)
∴0.5n48=188
∴n48=376
→n1=1.5n48=1.5 × 376=564
n42=n41-4=n1-4 × 41
∴n42=n1-160=564-164
∴n42=400Hz
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