/*! 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 39 For a wave displacement amplitud... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 39

For a wave displacement amplitude is \(10^{-8} \mathrm{~m}\), density of air \(1.3 \mathrm{~kg} \mathrm{~m}^{-3}\), velocity in air \(340 \mathrm{~ms}^{-1}\), and frequency is \(2000 \mathrm{~Hz}\). The intensity of wave is (A) \(5.3 \times 10^{-4} \mathrm{~W} / \mathrm{m}^{-2}\) (B) \(5.3 \times 10^{-6} \mathrm{~W} / \mathrm{m}^{-2}\) (C) \(3.5 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{-2}\) (D) \(3.5 \times 10^{-6} \mathrm{~W} / \mathrm{m}^{-2}\)

Short Answer

Expert verified
The intensity of the wave is approximately \(5.3 \times 10^{-6} \mathrm{~W} / \mathrm{m}^{-2}\).

Step by step solution

01

Substitute the given values

Let's substitute the values of amplitude, density, velocity, and frequency in the intensity formula. Amplitude = \(10^{-8} \mathrm{~m}\) Density = \(1.3 \mathrm{~kg} \mathrm{~m}^{-3}\) Velocity = \(340 \mathrm{~ms}^{-1}\) Frequency = 2000 Hz
02

Calculate the angular frequency

To find the angular frequency, we can use the following relation: Angular frequency = 2蟺 脳 frequency Angular frequency = 2蟺 脳 2000 Hz = 4000蟺 rad/s
03

Calculate the intensity of the wave

Now, we can use the formula: Intensity = \(\frac{1}{2}\) 脳 (density) 脳 (velocity) 脳 (angular frequency)虏 脳 (amplitude)虏 Intensity = \(\frac{1}{2}\) 脳 \(1.3 \mathrm{~kg} \mathrm{~m}^{-3}\) 脳 \(340 \mathrm{~ms}^{-1}\) 脳 \((4000\pi \mathrm{~rad} \mathrm{~s}^{-1})^2\) 脳 \((10^{-8} \mathrm{~m})^2\) Intensity 鈮� \(5.33 \times 10^{-6} \mathrm{~W} / \mathrm{m}^{-2}\) Comparing the result with the given options, the intensity of the wave is close to option (B): (B) \(5.3 \times 10^{-6} \mathrm{~W} / \mathrm{m}^{-2}\)

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Most popular questions from this chapter

Column-I (A) Beats (B) Standing waves (C) Interference (D) Echo Column-II 1\. Redistribution of energy 2\. Multiple reflection 3\. Varying amplitude 4\. Reflection from a rigid support

A whistle producing sound waves of frequencies \(9500 \mathrm{~Hz}\) and above is approaching a stationary person with speed \(v \mathrm{~ms}^{-1}\). The velocity of sound in air is \(300 \mathrm{~ms}^{-1}\). If the person can hear frequencies upto a maximum of \(10,000 \mathrm{~Hz}\), the maximum value of \(\mathrm{v}\) upto which he can hear whistle is (A) \(15 \sqrt{2} \mathrm{~ms}^{-1}\) (B) \(\frac{15}{\sqrt{2}} \mathrm{~ms}^{-1}\) (C) \(15 \mathrm{~ms}^{-1}\) (D) \(30 \mathrm{~ms}^{-1}\)

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