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 the preceding one. The
frequency of the first fork is 1.5 times the fre-
quency of the last fork. Find the frequency of the
first and the 42nd tuning fork. S)

Answer 1

Answer:

Give that

n₁ = 1.5n48 beat frequency = 4Hz

The set of turning forks are arranged in decreasing order of frequencies

.. n₂ = n1 - 4: n3 = n₂ - 4

= (n − 4) - 4

n48 1474= n₂ - 47 x 4

n48 n, - 188

n48 1.5n48 168..n₁ = 1.5n48

.. n48= 376

n₁ = 1.5n48 = 1.5 x 376 564

n42 = n41 -4 = n₁ + n x 41

.. n42 n₁160564 - 764

n42 400Hz