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Chapter 10: Problem 1508

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 fundamental mode of the closed pipe is $\ldots \ldots \ldots . h z$ (A) 100 (B) 200 (C) 300 (D) 400

Short Answer

Expert verified
The frequency of the fundamental mode of the closed pipe is \(200 \mathrm{~Hz}\).

Step by step solution

01

Calculate the speed of the wave on the string

First, we need to determine the linear mass density of the string, which is given by the mass divided by the length. Convert the mass and length to SI units: mass = 0.0025 kg and length = 0.25 m. Linear mass density (µ) = \( 0.0025/0.25 = 0.01 \mathrm{~kg/m} \).
02

Find the speed of the wave on the string using its overtone frequency

The string vibrates in its first overtone, which means the fundamental frequency is doubled. Let's denote the overtone frequency as \(2f_s\). The closed pipe has fundamental frequency \(f_p\), and the beat frequency is 8 Hz. So, \( |2f_s - f_p| = 8 \hspace{5 pt}\mathrm{Hz} \). Using the fact that the beat frequency decreases when the tension decreases, we can infer that the string's frequency is higher than the pipe's frequency. Therefore, \(2f_s - f_p = 8\).
03

Calculate the speed of the wave on the string using the closed pipe's fundamental frequency

For a closed pipe, the fundamental frequency is given by \(f_p = \cfrac{v}{4L}\), where v is the speed of sound and L is the length of the closed pipe, which is 0.4 m in this case. Therefore, \[f_p = \cfrac{320}{4 \cdot 0.4} = 200 \mathrm{~Hz} \]
04

Find the frequency of the string

Now, we can calculate the frequency of the string using the equation from Step 2: \(2f_s - 200 = 8\). Solving for the string's frequency, we get \(f_s = 104 \hspace{5 pt}\mathrm{Hz} \).
05

Calculate the speed and tension of the wave on the string

The speed of the wave on the string, v_s, can be calculated using the fundamental frequency, f_s, and the wavelength, which is equal to double the string's length: \(v_s = f_s \cdot \lambda = 104 \cdot 2 \cdot 0.25 = 52 \mathrm{~m/s} \). Finally, we can calculate the tension T, using the formula \(v_s = \sqrt{T/\mu}\). Thus, \(T= v_s^2\mu=52^2 \cdot 0.01 = 27.040 \mathrm{~N} \). So, the frequency of the fundamental mode of the closed pipe is \(200 \mathrm{~Hz}\), which corresponds to the option (B).

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Most popular questions from this chapter

A small spherical steel ball is placed at a distance slightly away from the center of a concave mirror having radius of curvature \(250 \mathrm{~cm}\). If the ball is released, it will now move on the curved surface. What will be the periodic time of this motion? Ignore frictional force and take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$. (A) \((\pi / 4) \mathrm{s}\) (B) \(\pi \mathrm{s}\) (C) \((\pi / 2) \mathrm{s}\) (D) \(2 \pi \mathrm{s}\)

The distance travelled by a particle performing S.H.M. during time interval equal to its periodic time is \(\ldots \ldots\) (A) A (B) \(2 \mathrm{~A}\) (C) \(4 \mathrm{~A}\) (D) Zero.

A block having mass \(\mathrm{M}\) is placed on a horizontal frictionless surface. This mass is attached to one end of a spring having force constant \(\mathrm{k}\). The other end of the spring is attached to a rigid wall. This system consisting of spring and mass \(\mathrm{M}\) is executing SHM with amplitude \(\mathrm{A}\) and frequency \(\mathrm{f}\). When the block is passing through the mid-point of its path of motion, a body of mass \(\mathrm{m}\) is placed on top of it, as a result of which its amplitude and frequency changes to \(\mathrm{A}^{\prime}\) and \(\mathrm{f}\). The ratio of amplitudes $\left(\mathrm{A}^{1} / \mathrm{A}\right)=\ldots \ldots \ldots$ (A) \(\sqrt{\\{}(\mathrm{M}+\mathrm{m}) / \mathrm{m}\\}\) (B) \(\sqrt{\\{m} /(\mathrm{M}+\mathrm{m})\\}\) (C) \(\sqrt{\\{} \mathrm{M} /(\mathrm{M}+\mathrm{m})\\}\) (D) \(\sqrt{\\{}(\mathrm{M}+\mathrm{m}) / \mathrm{M}\\}\)

For a particle executing \(\mathrm{S} . \mathrm{H} \mathrm{M} .\), when the potential energy of the oscillator becomes \(1 / 8\) the maximum potential energy, the displacement of the oscillator in terms of amplitude A will be........... (A) \((\mathrm{A} / \sqrt{2})\) (B) \(\\{\mathrm{A} /(2 \sqrt{2})\\}\) (C) \((\mathrm{A} / 2)\) (D) \(\\{\mathrm{A} /(3 \sqrt{2})\\}\)

If \((1 / 4)\) of a spring having length \(\ell\) is cutoff, then what will be the spring constant of remaining part? (A) \(\mathrm{k}\) (B) \(4 \mathrm{k}\) (C) \((4 \mathrm{k} / 3)\) (D) \((3 \mathrm{k} / 4)\)
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