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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

As shown in figure, a block A having mass \(M\) is attached to one end of a massless spring. The block is on a frictionless horizontal surface and the free end of the spring is attached to a wall. Another block B having mass ' \(\mathrm{m}\) ' is placed on top of block A. Now on displacing this system horizontally and released, it executes S.H.M. What should be the maximum amplitude of oscillation so that B does not slide off A? Coefficient of static friction between the surfaces of the block's is \(\mu\). (A) \(A_{\max }=\\{(\mu \mathrm{mg}) / \mathrm{k}\\}\) (B) \(A_{\max }=[\\{\mu(m+M) g\\} / k]\) (C) \(A_{\max }=[\\{\mu(M-\mathrm{m}) g\\} / \mathrm{k}]\) (D) \(A_{\max }=[\\{2 \mu(M+m)\\} / k]\)

Sound waves propagates with a speed of \(350 \mathrm{~m} / \mathrm{s}\) through air and with a speed of \(3500 \mathrm{~m} / \mathrm{s}\) through brass. If a sound wave having frequency \(700 \mathrm{~Hz}\) passes from air to brass, then its wavelength ......... (A) decreases by a fraction of 10 (B) increases 20 times (C) increases 10 times (D) decreases by a fraction of 20

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