/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 139 Sound waves of wavelength \(\lam... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 139

Sound waves of wavelength \(\lambda\) travelling in a medium with a speed of \(v \mathrm{~m} / \mathrm{s}\) enter into another medium where its speed in \(2 v \mathrm{~m} / \mathrm{s}\). Wavelength of sound waves in the second medium is (A) \(\lambda\) (B) \(\frac{\lambda}{2}\) (C) \(2 \lambda\) (D) \(4 \lambda\)

Short Answer

Expert verified
The short answer is: In the second medium, the new wavelength of sound waves is twice the initial wavelength. Therefore, the correct answer is (C) \(2 \lambda\).

Step by step solution

01

Understand the given information

The given information includes the initial wavelength \(\lambda\), initial speed \(v\), and the speed in the second medium \(2v\). We need to determine the wavelength of sound waves in the second medium.
02

Use the formula for the speed of sound waves

The formula relating the speed (\(v\)), frequency (\(f\)), and wavelength (\(\lambda\)) of a sound wave is given by: \[v = f\lambda\]
03

Determine the frequency of the sound wave in the first medium

Using the given information, we can find the frequency of the sound wave in the first medium, using the formula \(v = f\lambda\): \[f = \frac{v}{\lambda}\]
04

Use the fact that frequency is constant in different media

The frequency of the sound wave does not change between different media, so the frequency we found in the first medium also applies to the second medium. In other words: \[f_{2} = f_{1}\]
05

Determine the wavelength in the second medium

Now we know the frequency and speed of the sound wave in the second medium (\(f_{2} = f_{1}\) and \(v_{2} = 2v_{1}\)). We can use the formula for speed from Step 2 to find the wavelength in the second medium: \[v_{2} = f_{2}\lambda_{2}\] Substitute, using \(v_{2} = 2v_{1}\) and \(f_{2} = \frac{v_{1}}{\lambda_{1}}\): \[2v_{1} = (\frac{v_{1}}{\lambda_{1}}) \cdot \lambda_{2}\] \[2 = (\frac{1}{\lambda_{1}}) \cdot \lambda_{2}\] \[\lambda_{2} = 2\lambda_{1}\] So, the new wavelength in the second medium is twice the initial wavelength. The correct answer is (C) \(2 \lambda\).

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