# 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. - Physics

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

Concept: Formation of Beats
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