Question

A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamentals frequency of the pipe if the tension in the wire is 50 N and the speed of sound is 320 ms^–1, what is the mass of the string. 

(A) 20 g 

(B) 5 g 

(C) 40 g 

(D) 10 g

Answer 1

The correct option is (D) 10 g.

Explanation:

ℓ = length of string = 0.5 m 

L = length of pipe = 0.8 m 

frequency of string vibrating in the second harmonics is 

V = (2 / 2ℓ) √(T/μ) Here μ is mass per unit length of string. 

Fundamental frequency of closed pipe is 

V' = {Ʋ / 4L} where Ʋ is speed of sound 

For resonance V = V' hence 

(2 / 2ℓ) √(T/μ) = (Ʋ / 4L) 

∴ {1 / (0.5)} √(50 / μ) = {(320) / (4 × 0.8)} 

∴ √(50 / μ) = {(160) / (4 × 0.8)} 

∴ (50 / μ) = 2500 

∴ μ = {50 / (2500)} = (1 / 50) 

∴ mass of string = m = μℓ = (1 / 50) × 0.5 = {1 / (100)}kg = 10 gm