Question

When a vibrating fork is placed on a sound box of a sonometer. 8 beats per second are heard when the length of the sonometer wire is kept at 101 cm or 100 cm. Then the frequency of the tuning fork is (consider that the tension in the wire is kept constant).:
A. $1616Hz$
B. $1608Hz$
C. $1632Hz$
D. $1600Hz$

Answer 1

There are certain things we know, like absolute difference of two different frequencies produce beats. The multiplication of frequency and length of wire Is constant, that is “The frequency produced by wire is inversely proportional to the length of the wire”. As another quantity increases another quantity decreases thus their product remains constant.

Complete step-by-step solution:
As we know that
$n \times l$ is always constant.
Therefore, for two different situations:
${n_1}{l_1} = {n_2}{l_2}$
So we can say that:
In first case: $(n + 8) \times 100$ = constant
In second case $(n - 8) \times 101$ = constant
Constant in both cases are equal therefore::
$(n + 8) \times 100 = (n - 8) \times 101$
$\dfrac{{n + 8}}{{n - 8}} = \dfrac{{101}}{{100}}$
$(101 - 100)n = 800 + 808$
$n = 1608Hz$
Hence option B is correct.

Note:-
Relationship between frequency and several quantities:
1. In a vibrating string the fundamental frequency of is inversely proportional to its length.
2. In a vibrating string the fundamental frequency of is directly proportional to the square root of the tension.
3. In a vibrating string the fundamental frequency of is inversely proportional to the square root of the mass per unit length.