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 resonated with the fundamental frequency of the pipe. If the tension in the wire is 50N and the speed of sound is 320m/s, the mass of the string is:
A. 5 grams
B. 10 grams
C. 20 grams
D. 40 grams

Answer 1

Hint: A transverse wave is the one which travels through a medium, making crests and troughs. Wave on a string is an example of a transverse wave. A longitudinal wave is the one which travels through the medium making compressions and rarefactions. Sound waves are an example of longitudinal waves. When two waves approach each other in the same place, on the same line, standing waves are formed.
Formula used:
$\nu = \dfrac{v}{4L} (for\ closed\ organ\ pipe)\ and\ frequency_{2nd Harmonic} = \dfrac{2}{2l}\sqrt{\dfrac{T}{\mu}}$

Complete answer:
Organ pipe is related to the experiments on sound waves. Open organ pipe is the one in which both ends are opened and then sound is passed through it. Closed organ pipe is the one in which only one end is open and the other is closed and then sound is passed.
Now, for a closed organ pipe, the fundamental frequency is given$\nu = \dfrac{v}{4L}$, where ‘v’ is the velocity of sound in the medium of organ pipe and ‘L’ being the length of pipe.
As the pipe is closed, thus we can apply the formulas for closed organ pipe on it.
Now, we are given second harmonic frequency of wire is equal and in resonance with fundamental frequency of pipe, thus;
$\nu = \dfrac{v}{4L} = \dfrac{1}{l}\sqrt{\dfrac{T}{\mu}}$
Putting T = 50 N, v = 320 m/s and l = 0.8 m.
$\dfrac{320}{4(0.8)} = \dfrac{1}{0.5}\sqrt{\dfrac{50}{\mu}}$
$\implies \sqrt{\dfrac{\mu}{50} } = \dfrac{1}{50}$
Or $\mu = 0.02 kg/m$
Now, length of string = 0.5 m
Thus, mass of string = $0.02\times 0.5 = 0.01 kg = 10 g$.

So, the correct answer is “Option B”.

Note:
One must know that sound waves create standing waves inside the tube so that we can analyze the situation and know the parameters like frequency and conduct the experiment. Students are advised to derive the expression of harmonic frequencies once by their own.