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 frequency of the string vibrating in its \(1^{\text {st }}\) overtone is \(\ldots \ldots \ldots . \mathrm{Hz}\) (A) 92 (B) 108 (C) 192 (D) 208 .

Short answer

The frequency of the string vibrating in its \(1^{\text{st}}\) overtone is \(208 \mathrm{~Hz}\) (Answer: D).

Step-by-step solution

Understand and write down the given information

From the exercise, we have the following information: Length of the string (L) \(= 25\mathrm{~cm}\) mass of the string (m) \(= 2.5\mathrm{~g}\) Length of the closed pipe \(= 40\mathrm{~cm}\) Speed of sound in air \(= 320\mathrm{~ms}^{-1}\) Beat frequency \(= 8\mathrm{~Hz}\)

Calculate the fundamental frequency of the closed pipe

Since the pipe is closed at one end, its fundamental frequency is given by: \(f_{\text{pipe}} = \frac{v}{4 \times L_{\text{pipe}}}\) Where \(v\) is the speed of sound and \(L_{\text{pipe}}\) is the pipe's length. Plugging in the given values, \(f_{\text{pipe}} = \frac{320\mathrm{~ms}^{-1}}{4\times0.40\mathrm{~m}} = 200\mathrm{~Hz}\)

Calculate the linear density of the vibrating string

To calculate the frequency of the string in its first overtone, we will need its linear density, which is given by: \(\mu = \frac{m}{L}\) Converting the mass and length into SI units, we have: \(m\) \(= 2.5\mathrm{~g} = 0.0025\mathrm{~kg}\) \(L\) \(= 25\mathrm{~cm} = 0.25\mathrm{~m}\) Now, calculating the linear density: \(\mu = \frac{0.0025\mathrm{~kg}}{0.25\mathrm{~m}} = 0.01\mathrm{~kg/m}\)

Calculate the string's first overtone frequency

Now, considering the beat frequency, we have that the string's first overtone frequency can either be 8 Hz higher or lower than the fundamental frequency of the pipe: \(f_1 = f_{\text{pipe}} + 8\mathrm{~Hz} = 200\mathrm{~Hz} + 8\mathrm{~Hz} = 208\mathrm{~Hz}\) \(f_2 = f_{\text{pipe}} - 8\mathrm{~Hz} = 200\mathrm{~Hz} - 8\mathrm{~Hz} = 192\mathrm{~Hz}\) Next, we'll apply the formula for the frequency of the string in the first overtone mode: \(f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\) Since decreasing the tension in the string decreases the beat frequency, we should choose \(f_1 = 208 \mathrm{~Hz}\) as the frequency for the first overtone. Therefore, the frequency of the string vibrating in its \(1^{\text{st}}\) overtone is \(208 \mathrm{~Hz}\) (Answer: D).