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Chapter 10: Problem 1485

A vehicle with a horn of frequency \(\mathrm{n}\) is moving with a velocity of \(30 \mathrm{~m} / \mathrm{s}\) in a direction perpendicular to the straight line joining the observer and the vehicle. The observer perceives the sound to have a frequency \(\left(\mathrm{n}+\mathrm{n}_{1}\right) .\) If the sound velocity in air is \(300 \mathrm{~m} / \mathrm{s}\), then \(\ldots \ldots . .\) (A) \(\mathrm{n}_{1}=10 \mathrm{n}\) (B) \(\mathrm{n}_{1}=0\) (C) \(\mathrm{n}_{1}=0.1 \mathrm{n}\) (D) \(\mathrm{n}_{1}=-0 . \ln\)

Short Answer

Expert verified
The correct answer is (C) \(\mathrm{n}_{1} = 0.1 \mathrm{n}\).

Step by step solution

01

Understanding the Doppler effect formula

The Doppler effect formula relates the observed frequency (f' ) to the source frequency (f), the speed of sound (v), the speed of the source (vs), and the speed of the observer (vo). The formula is given as follows: \(f'=f\left(\frac{v\pm vo}{v\mp vs}\right)\) Here, f' = observed frequency, f = source frequency, v = speed of sound, vo = speed of the observer, and vs = speed of the source. In this case, the source is the vehicle and the observer is stationary. Hence, vo = 0 m/s.
02

Apply the given values to the Doppler effect formula

We are given: n = source frequency, n + n1 = observed frequency, v = 300 m/s (speed of sound), and vs = 30 m/s (speed of the source, the vehicle). Since the vehicle is moving perpendicular to the observer and the observer is stationary, the Doppler effect formula becomes: \((\mathrm{n}+\mathrm{n}_{1})=n\left(\frac{v}{v-vs}\right)\)
03

Solve for n1

Now we have: \((\mathrm{n}+\mathrm{n}_{1})=n\left(\frac{300}{300-30}\right)\) Divide both sides by n: \(\frac{(\mathrm{n}+\mathrm{n}_{1})}{n} = \frac{300}{(300-30)}\) \((1+\frac{\mathrm{n}_{1}}{\mathrm{n}})=\frac{300}{270}\) \(\frac{\mathrm{n}_{1}}{\mathrm{n}}=\frac{300}{270}-1\) \(\frac{\mathrm{n}_{1}}{\mathrm{n}}=0.1\)
04

Compare with the given options

We are to find which of the given options is correct: (A) \(\mathrm{n}_{1} = 10 \mathrm{n}\) (B) \(\mathrm{n}_{1} = 0\) (C) \(\mathrm{n}_{1} = 0.1 \mathrm{n}\) (D) \(\mathrm{n}_{1} = -0 . \ln\) Comparing the answer, \(\frac{\mathrm{n}_{1}}{\mathrm{n}} = 0.1\), with the given options, we can see that the correct answer is (C) \(\mathrm{n}_{1} = 0.1 \mathrm{n}\).

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Most popular questions from this chapter

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) \(\\{\mathrm{A} /(3 \sqrt{2})\\}\)

The speed of a particle executing motion changes with time according to the equation $\mathrm{y}=\mathrm{a} \sin \omega \mathrm{t}+\mathrm{b} \cos \omega \mathrm{t}\(, then \)\ldots \ldots \ldots$ (A) Motion is periodic but not a S.H.M. (B) It is a S.H.M. with amplitude equal to \(\mathrm{a}+\mathrm{b}\) (C) It is a S.H.M. with amplitude equal to \(\mathrm{a}^{2}+\mathrm{b}^{2}\)

Two wires made up of same material are of equal lengths but their radii are in the ratio \(1: 2\). On stretching each of these two strings by the same tension, the ratio between their fundamental frequency is \(\ldots \ldots \ldots .\) (A) \(1: 2\) (B) \(2: 1\) (C) \(1: 4\) (D) \(4: 1\)

The function \(\sin ^{2}(\omega t)\) represents (A) A SHM with periodic time \(\pi / \omega\) (B) A SHM with a periodic time \(2 \pi / \omega\) (C) A periodic motion with periodic time \(\pi / \omega\) (D) A periodic motion with period \(2 \pi / \omega\)

A particle having mass \(1 \mathrm{~kg}\) is executing S.H.M. with an amplitude of \(0.01 \mathrm{~m}\) and a frequency of \(60 \mathrm{hz}\). The maximum force acting on this particle is \(\ldots \ldots . . \mathrm{N}\) (A) \(144 \pi^{2}\) (B) \(288 \pi^{2}\) (C) \(188 \pi^{2}\) (D) None of these. (A) \(x=a \sin 2 p \sqrt{(\ell / g) t}\) (B) \(x=a \cos 2 p \sqrt{(g / \ell) t}\) (C) \(\mathrm{x}=\mathrm{a} \sin \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\) (D) \(\mathrm{x}=\mathrm{a} \cos \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\)
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