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# MCQs of Beats

Question:
56 tuning forks are so arranged in series, that each fork gives 4 beats/second with the previous one. If the frequency of the last fork is 3 times that of the first, then the frequency of the first fork will be
a) 55 Hz
b) 110 Hz
c) 75 Hz
d) 220 Hz

Solution:
In such sort of problems you need to form the equations from the given data which can be bit troublesome for students of pre medical. If you follow the steps, question wont be hard.

Total number of forks= n = 56
Let the frequency of first fork be x and the last fork be y.
According to given data frequency of last for is 3 times the first one, so we can write:
y = 3x

Each fork gives 4 beats with the previous one, which means there is a difference of 4 Hz in frequencies of two successive forks. So, if frequency of first fork is x, frequency of 2nd fork will be x + 4, for 3rd it will be x + 4 + 4 = x + 8, for 4th it will be x + 8 + 4 = x + 12 and so on.

So we get the following series:
x , x+4 , x+8, x+12, x+16, ...., y

It can also be written as:
x + 0(4) , x + 1(4), x + 2(4),  x + 3(4), .... , y

x + 0(4)= x + (1-1) (4) =  frequency of 1st fork
x + 1(4) =x + (2-1)(4)  = frequency of 2nd fork

So this way frequency of 56th fork will be:

y = x + (56-1)(4) = x + 55(4) = x +  220

So we got two equations:
y = 3x  --- (i)
y = x + 220 --- (ii)

Use value of y from (i) in (ii). We'll get:
3x = x + 220
2x = 220
x = 110 Hz = Frequency of first fork.

I Hope its clear now..