Question

A stretched sonometer wire is in unison with a tuning fork. when the length of the wire is increased by 2% the number of beats heard per scond is 5 then, the frequency of the fork is

Answer 1

Let the Length of wire is L
Length is increased by 2 % then number of beats heard per second is 5
Means n₁ - n₂ = 5sec⁻¹ or, 5Hz -----(1)
Here n₁ and n₂ are the frequencies of intial and final.

we know, n = $\frac{1}{2l}\sqrt{\frac{T}{m}}$ , here n is frequency , l is the length of wire , m is the mass per unit length and T is the tension in string.
so, here it is clear that frequency is inversely proportional to length of wire.
∴n₁l₁ = n₂l₂
n₁L = n₂1.02L { wire is increased by 2%, so L₂ = 1.02L}
50n₁ = 51n₂
n₂ = 50/51n₁ ----------(2)

from equations (1) and (2)
n₁ - 50/51n₁ = 5
⇒(51 - 50)/51 n₁ = 5
⇒ 1/51 n₁ = 5
⇒n₁ = 51 × 5 Hz = 255Hz

Hence , frequency of turning fork is 255Hz