Question

A string 25 cm long and having mass of 2.5 g is under tension. A pipe closed at one end is 40 cm long. When the string is set vibration in its first overtone and the air in the pipe in its fundamental frequency, 8 beats per second are heard. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is 320 m/s, then the tension in the string is\[\left( {20 + x} \right)N\]. Find x.

Answer 1

Hint: It is given that a closed organ pipe is 40cm long and a stretched string is under a certain tension. Longitudinal vibrations of this stretched string can be set up in the closed organ pipe. The beats heard per second is the frequency of the closed organ pipe subtracted from the frequency of the \[{m^{th}}\] harmonic of the string.

Formula used:
The fundamental frequency of closed organ pipe is given by: \[\dfrac{v}{{4L}}\]
The frequency of the \[{m^{th}}\] harmonic or the \[{(m - 1)^{th}}\] overtone of the string is given by: \[{n_m} = \dfrac{1}{l}\sqrt {\dfrac{T}{\mu }} \]

Complete step by step answer:
A string is a thin flexible thread or wire. Transverse stationary vibration can be set up in a stretched string. On the other hand, an organ pipe is a hollow tube having length much greater than it’s diameter. Longitudinal stationary vibrations can be set up in an organ pipe. In an organ pipe, if both the ends are open, then it is an open organ pipe. If one end is closed, then it is a closed organ pipe.
In the problem, it is given that
Length of the string=\[l = 25cm = 0.25m\]
Mass of the string=\[m = 2.5g = 0.0025kg\]
Length of the organ pipe=\[L = 40cm = 0.40m\]
The frequency of the \[{m^{th}}\] harmonic or the \[{(m - 1)^{th}}\] overtone of the string is given by
\[{n_m} = \dfrac{1}{l}\sqrt {\dfrac{T}{\mu }} \]\[ \to \](1)
where \[T\]is the tension in the string and \[\mu \]is the mass per unit length of the string.
Therefore, \[\mu \]will be the mass of the string divided by the per unit length of the string.
\[\mu = \dfrac{{0.0025}}{{0.25}} = 0.01\]
\[\therefore {n_m} = \dfrac{1}{l}\sqrt {\dfrac{T}{\mu }} = \dfrac{1}{{0.25}}\sqrt {\dfrac{T}{{0.01}}} = 40\sqrt T \]
The fundamental frequency of a closed organ pipe is given by \[\dfrac{v}{{4L}}\]. Thus,
\[\dfrac{v}{{4L}} = \dfrac{{320}}{{(4 \times 0.40)}} = 200Hz\]
where \[v\]is the speed of sound in air. In the problem it is given by 320 m/s.
The beat frequency is given as 8
\[\therefore 40\sqrt T - 200 = 8\]
$\Rightarrow \sqrt T = \dfrac{{208}}{{40}} = 5.2 \\
  \therefore T = 27.04N \\ $
In the problem, the tension in the string is given as\[\left( {20 + x} \right)N\]. Equating the two sides of the equation to find \[x\]
\[ T = 27.04N = \left( {20 + x} \right)N\]
$\Rightarrow 27.04N = \left( {20 + x} \right)N \\
\Rightarrow x = 27.04 - 20 \\
\therefore x = 7.04 \\ $

Note:
Universally the fundamental frequency of a closed organ pipe is 220Hz.
The mode of vibration always depends on the position from where the string is plucked. For example if the string is plucked from the middle, then the string vibrates in the fundamental mode producing the first harmonic.