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Chapter 10: Problem 1492

A car blowing its horn at \(480 \mathrm{~Hz}\) moves towards a high wall at a speed of \(20 \mathrm{~m} / \mathrm{s}\). If the speed of sound is $340 \mathrm{~m} / \mathrm{s}$, the frequency of the reflected sound heard by the driver sitting in the car will be closest to $\ldots \ldots \ldots . \mathrm{Hz}$ (A) 510 (B) 524 (C) 568 (D) 480

Short Answer

Expert verified
The frequency of the reflected sound heard by the driver sitting in the car is closest to \(524 \mathrm{~Hz}\).

Step by step solution

01

List given values

Let's first list the given values: - Original frequency of car horn, \(f_0 = 480 \mathrm{~Hz}\) - Speed of car, \(v_\mathrm{car} = 20 \mathrm{~m/s}\) - Speed of sound, \(v_\mathrm{sound} = 340 \mathrm{~m/s}\)
02

Calculate the frequency of the sound that bounces off the wall

Because the car and driver are moving towards the wall, we will use the Doppler effect formula for the frequency observed by the wall: \(f_\mathrm{wall} = \frac{v_\mathrm{sound} + v_\mathrm{wall}}{v_\mathrm{sound} - v_\mathrm{car}} \cdot f_0\) Since the wall is stationary, \(v_\mathrm{wall} = 0\), and we can simplify the formula as: \(f_\mathrm{wall} = \frac{v_\mathrm{sound}}{v_\mathrm{sound} - v_\mathrm{car}} \cdot f_0\) Now, substitute the given values and calculate \(f_\mathrm{wall}\): \(f_\mathrm{wall} = \frac{340}{340 - 20} \cdot 480\) \(f_\mathrm{wall} = \frac{340}{320} \cdot 480\) \(f_\mathrm{wall} = 1.0625 \cdot 480 = 510 \mathrm{~Hz}\)
03

Calculate the frequency of the reflected sound heard by the driver

Lastly, we need to calculate the frequency of the reflected sound as observed by the driver who is now considered an observer moving towards the wall and source. We will use the Doppler effect formula again, but this time with \(f_\mathrm{wall}\) as the source frequency and taking into account the movement of the observer: \(f_\mathrm{driver} = \frac{v_\mathrm{sound} + v_\mathrm{car}}{v_\mathrm{sound}} \cdot f_\mathrm{wall}\) Now, substitute the given values and calculate \(f_\mathrm{driver}\): \(f_\mathrm{driver} = \frac{340 + 20}{340} \cdot 510\) \(f_\mathrm{driver} = \frac{360}{340} \cdot 510\) \(f_\mathrm{driver} = 1.0588 \cdot 510 = 524 \mathrm{~Hz}\) (rounded to the nearest whole number) So, the frequency of the reflected sound heard by the driver sitting in the car is closest to \(524 \mathrm{~Hz}\), which corresponds to option (B).

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Most popular questions from this chapter

When an elastic spring is given a displacement of \(10 \mathrm{~mm}\), it gains an potential energy equal to \(\mathrm{U}\). If this spring is given an additional displacement of \(10 \mathrm{~mm}\), then its potential energy will be.............. (A) \(\mathrm{U}\) (B) \(2 \mathrm{U}\) (C) \(4 \mathrm{U}\) (D) \(\mathrm{U} / 4\).

The equation for displacement of a particle at time \(t\) is given by the equation \(\mathrm{y}=3 \cos 2 \mathrm{t}+4 \sin 2 \mathrm{t}\). The maximum acceleration of the particle is $\ldots \ldots . . \mathrm{cm} / \mathrm{s}^{2}$. (A) 4 (B) 12 (C) 20 (D) 28

Which of the equation given below represents a S.H.M.? (A) acceleration \(=-\mathrm{k}(\mathrm{x}+\mathrm{a})\) (B) acceleration \(=\mathrm{k}(\mathrm{x}+\mathrm{a})\) (C) acceleration \(=\mathrm{kx}\) (D) acceleration \(=-\mathrm{k}_{0} \mathrm{x}+\mathrm{k}_{1} \mathrm{x}^{2}\) \\{Here \(\mathrm{k}, \mathrm{k}_{0}\) and \(\mathrm{k}_{1}\) are force constants and units of \(\mathrm{x}\) and a is meter \(\\}\)

As shown in figure, a spring attached to the ground vertically has a horizontal massless plate with a \(2 \mathrm{~kg}\) mass in it. When the spring (massless) is pressed slightly and released, the \(2 \mathrm{~kg}\) mass, starts executing S.H.M. The force constant of the spring is \(200 \mathrm{Nm}^{-1}\). For what minimum value of amplitude, will the mass loose contact with the plate? (Take \(\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\) (A) \(10.0 \mathrm{~cm}\) (B) \(8.0 \mathrm{~cm}\) (C) \(4.0 \mathrm{~cm}\) (D) For any value less than \(12.0 \mathrm{~cm}\).

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) \(A_{\max }=[\\{2 \mu(M+m)\\} / k]\)
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