Question

A string \(25 \mathrm{~cm}\) long and having a mass of \(2.5 \mathrm{~g}\) is under tension. A pipe closed at one end is \(40 \mathrm{~cm}\) long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats per second is heard. It is observed that decreasing the tension in the string decreases the beat frequency. The speed of sound in air is \(320 \mathrm{~ms}^{-1}\) The tension in the string is very nearly equal to \(\ldots \ldots \ldots\) (A) \(25 \mathrm{~N}\) (B) \(27 \mathrm{~N}\) (C) \(28 \mathrm{~N}\) (D) \(30 \mathrm{~N}\)

Short answer

The tension in the string is very nearly equal to \(27 \mathrm{~N}\) (B).

Step-by-step solution

Find the frequency of the first overtone of the string

To compute the frequency of the first overtone of the string, we must first recall the formula for the fundamental frequency of a string, which is given by \(f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\), where f1 is the fundamental frequency, L is the length of the string, T is the tension, and μ is the linear mass density of the string. Since the string is vibrating in its first overtone, the frequency of this overtone will be twice the fundamental frequency, which is given by \(f_2 = 2f_1 = \frac{1}{L} \sqrt{\frac{T}{\mu}}\).

Find the linear mass density of the string

Let's find the linear mass density (μ) of the string, which is given by the mass divided by the length of the string. In this case, the mass 𝑚 is 2.5g and the length 𝐿 is 25cm. The linear mass density μ can be calculated as follows: \(\mu = \frac{m}{L} = \frac{2.5 \times 10^{-3} \mathrm{~kg}}{25 \times 10^{-2} \mathrm{~m}} = 10^{-4} \mathrm{kg/m}\)

Calculate the fundamental frequency of the closed pipe

The formula for the fundamental frequency of a closed pipe is given by \(f_p = \frac{v}{4L_p}\), where \(f_p\) represents the fundamental frequency of the pipe, \(v\) is the speed of sound in air, and \(L_p\) is the length of the pipe. Using the given values for the speed of sound (320 m/s) and the pipe length (40 cm), we can find the fundamental frequency: \(f_p = \frac{320 \mathrm{~ms^{-1}}}{4 \times 0.4 \mathrm{~m}} = 200 \mathrm{Hz}\)

Find the frequency of the first overtone of the string

Let's now find the frequency of the first overtone of the string (\(f_2\)). Since we know the beat frequency (8 beats/s) and that decreasing the tension in the string decreases the beat frequency, this suggests that the frequency of the first overtone of the string is greater than the fundamental frequency of the closed pipe: \(f_2 = f_p + 8 \mathrm{~Hz} = 200 \mathrm{~Hz} + 8 \mathrm{~Hz} = 208 \mathrm{~Hz}\)

Calculate the tension in the string

Now that we have the frequency of the first overtone of the string, the linear mass density of the string, and the length of the string, we can solve for the tension in the string using the formula for the first overtone frequency: \(f_2 = \frac{1}{L} \sqrt{\frac{T}{\mu}}\) Rearranging this equation to solve for the tension, we get: \(\ T = \mu L^2 f_2^2\) Plugging in the values, we can find the tension in the string: \(\ T = (10^{-4} \mathrm{kg/m}) (0.25 \mathrm{~m})^2 (208 \mathrm{Hz})^2 \) \(T \approx 27 \mathrm{~N}\) So, the tension in the string is very nearly equal to 27 N, which corresponds to option (B).