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Chapter 10: Problem 1446

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

Short Answer

Expert verified
The wavelength of the sound wave increases 10 times as it passes from air to brass.

Step by step solution

01

Determine the wavelength in air

According to the exercise, the speed of sound in air is 350 m/s, and the frequency of the sound wave is 700 Hz. We'll start by finding the wavelength in air using the formula: Wave speed = Frequency × Wavelength \(350 = 700 \times \lambda_\mathrm{air}\) Divide both sides by 700 to find the wavelength in air: \(\lambda_\mathrm{air} = \frac{350}{700}\) \(\lambda_\mathrm{air} = 0.5 \mathrm{~m}\)
02

Determine the wavelength in brass

Now we'll find the wavelength in brass, using the same formula. The speed of sound in brass is 3500 m/s, and the frequency remains the same at 700 Hz: \(3500 = 700 \times \lambda_\mathrm{brass}\) Divide both sides by 700 to find the wavelength in brass: \(\lambda_\mathrm{brass} = \frac{3500}{700}\) \(\lambda_\mathrm{brass} = 5 \mathrm{~m}\)
03

Determine the change in wavelength

Now that we've found the wavelengths in air and brass, we can determine the change in wavelength: \(\frac{\lambda_\mathrm{brass}}{\lambda_\mathrm{air}} = \frac{5}{0.5}\) This simplifies to: \(\frac{\lambda_\mathrm{brass}}{\lambda_\mathrm{air}} = 10\) So, the wavelength in brass increases 10 times compared to the wavelength in air.
04

Match the result with answer choices

Now, we match our result with the answer choices provided. The wavelength increases 10 times, which matches with answer choice (C). Therefore, the correct answer is: (C) increases 10 times

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Most popular questions from this chapter

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) \(\theta+(\pi / 2)\).

A person standing in a stationary lift measures the periodic time of a simple pendulum inside the lift to be equal to \(\mathrm{T}\). Now, if the lift moves along the vertically upward direction with an acceleration of $(\mathrm{g} / 3)$, then the periodic time of the lift will now be \((\mathrm{A}) \sqrt{3} \mathrm{~T}\) (B) \(\sqrt{(3 / 2) \mathrm{T}}\) (C) \((\mathrm{T} / 3)\) (D) \((\mathrm{T} / \sqrt{3})\)

A particle executing S.H.M. has an amplitude \(\mathrm{A}\) and periodic time \(\mathrm{T}\). The minimum time required by the particle to get displaced by \((\mathrm{A} / \sqrt{2})\) from its equilibrium position is $\ldots \ldots \ldots \mathrm{s}$. (A) \(\mathrm{T}\) (B) \(\mathrm{T} / 4\) (C) \(\mathrm{T} / 8\) (D) \(\mathrm{T} / 16\)

Tube \(A\) has both ends open while tube \(B\) has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube \(\mathrm{A}\) and \(\mathrm{B}\) is \(\ldots \ldots \ldots\) (A) \(1: 2\) (B) \(1: 4\) (C) \(2: 1\) (D) \(4: 1\)

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) \(350 \mathrm{~m} / \mathrm{s}\)
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