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Chapter 10: Problem 1480

A closed organ pipe has fundamental frequency \(100 \mathrm{~Hz}\). What frequencies will be produced if its other end is also opened? (A) \(200,400,600,800 \ldots \ldots\) (B) \(200,300,400,500 \ldots \ldots\) (C) \(100,300,500,700 \ldots \ldots\) (D) \(100,200,300,400 \ldots \ldots\)

Short Answer

Expert verified
When the other end of the organ pipe is opened, the frequencies produced are 200 Hz, 400 Hz, 600 Hz, 800 Hz, and so on. The correct answer is (A).

Step by step solution

01

Understanding the formula for fundamental frequency

The fundamental frequency of a closed organ pipe is given by \(f_{closed}=\frac{v}{4L}\), where \(v\) is the speed of sound and \(L\) is the length of the pipe. For an open organ pipe, it is given by \(f_{open}=\frac{v}{2L}\).
02

Using the given information

The problem states that the fundamental frequency of theclosed organ pipe is 100 Hz. We can use this information to find the length of the pipe. From the formula, \(f_{closed}=\frac{v}{4L}\), we can write the equation: \(100 = \frac{v}{4L}\)
03

Find the fundamental frequency of open organ pipe

Now we can find the fundamental frequency of the open organ pipe using the formula: \(f_{open}=\frac{v}{2L}\). Since we have the equation \(100 = \frac{v}{4L}\), we can multiply both sides by 2 to get: \(200 = \frac{v}{2L}\). Thus, when opened, the organ pipe has a fundamental frequency of 200 Hz.
04

Find the higher order frequencies of open organ pipe

For an open organ pipe, the higher-order frequencies are given by the formula \(f_n = n\cdot f_{open}\), where \(n\) is an integer. Using the fundamental frequency of 200 Hz for the open organ pipe, we can find the next frequencies by multiplying by integers: \(f_2 = 2 \cdot 200 = 400~\text{Hz}\) \(f_3 = 3 \cdot 200 = 600~\text{Hz}\) \(f_4 = 4 \cdot 200 = 800~\text{Hz}\) Thus, the frequencies produced when the other end of the organ pipe is also opened are 200, 400, 600, 800... The correct answer is (A).

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Most popular questions from this chapter

If the mass of 1 mole of air is \(29 \times 10^{-3} \mathrm{~kg}\), then the speed of sound in it at STP is $(\gamma=7 / 5) .\left\\{\mathrm{T}=273 \mathrm{~K}, \mathrm{P}=1.01 \times 10^{5} \mathrm{~Pa}\right\\}$ (A) \(270 \mathrm{~m} / \mathrm{s}\) (B) \(290 \mathrm{~m} / \mathrm{s}\) (C) \(330 \mathrm{~m} / \mathrm{s}\) (D) \(350 \mathrm{~m} / \mathrm{s}\)

If the velocity of sound wave in humid air is \(\mathrm{v}_{\mathrm{m}}\) and that in dry air is \(\mathrm{v}_{\mathrm{d}}\), then \(\ldots \ldots\) (A) \(\mathrm{v}_{\mathrm{m}}>\mathrm{v}_{\mathrm{d}}\) (B) \(\mathrm{v}_{\mathrm{m}}<\mathrm{v}_{\mathrm{d}}\) (C) \(\mathrm{v}_{\mathrm{m}}=\mathrm{v}_{\mathrm{d}}\) \((\mathrm{D}) \mathrm{v}_{\mathrm{m}} \gg \mathrm{v}_{\mathrm{d}}\)

A body having mass \(5 \mathrm{~g}\) is executing S.H.M. with an amplitude of \(0.3 \mathrm{~m}\). If the periodic time of the system is $(\pi / 10) \mathrm{s}\(, then the maximum force acting on body is \)\ldots \ldots \ldots \ldots$ (A) \(0.6 \mathrm{~N}\) (B) \(0.3 \mathrm{~N}\) (C) \(6 \mathrm{~N}\) (D) \(3 \mathrm{~N}\)

If the equation for displacement of two particles executing S.H.M. is given by \(\mathrm{y}_{1}=2 \sin (10 \mathrm{t}+\theta)\) and $\mathrm{y}_{2}=3 \cos 10 \mathrm{t}$ respectively, then the phase difference between the velocity of two particles will be \(\ldots \ldots \ldots\) (A) \(-\theta\) (B) \(\theta\) (C) \(\theta-(\pi / 2)\) (D) \(\theta+(\pi / 2)\).

If two almost identical waves having frequencies \(\mathrm{n}_{1}\) and \(\mathrm{n}_{2}\), produced one after the other superposes then the time interval to obtain a beat of maximum intensity is \(\ldots \ldots \ldots .\) (A) \(\left\\{1 /\left(\mathrm{n}_{1}-\mathrm{n}_{2}\right)\right\\}\) (B) \(\left(1 / \mathrm{n}_{1}\right)-\left(1 / \mathrm{n}_{2}\right)\) (C) \(\left(1 / \mathrm{n}_{1}\right)+\left(1 / \mathrm{n}_{2}\right)\) (D) \(\left\\{1 /\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)\right\\}\)
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