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Chapter 10: Problem 1460

A wave travelling along a string is described by $\mathrm{y}=0.005 \sin (40 \mathrm{x}-2 \mathrm{t})$ in SI units. The wavelength and frequency of the wave are \(\ldots \ldots \ldots\) (A) \((\pi / 5) \mathrm{m} ; 0.12 \mathrm{~Hz}\) (B) \((\pi / 10) \mathrm{m} ; 0.24 \mathrm{~Hz}\) (C) \((\pi / 40) \mathrm{m} ; 0.48 \mathrm{~Hz}\) (D) \((\pi / 20) \mathrm{m} ; 0.32 \mathrm{~Hz}\)

Short Answer

Expert verified
The wavelength and frequency of the wave are approximately (D) $(\frac{\pi}{20})m; 0.32 Hz$.

Step by step solution

01

Identify the wave number (k) and angular frequency (w)

From the given wave equation: y = 0.005 * sin(40x - 2t) We can see that the wave number (k) is 40, and the angular frequency (w) is 2.
02

Find the wavelength (λ) from the wave number (k)

Using the relationship k = 2 * pi / λ, we can find the wavelength (λ) by rearranging the equation: λ = 2 * pi / k Plug in the value of k (40) to get: λ = \( \frac{2 * \pi}{40} \) λ = \( \frac{\pi}{20} \) meters
03

Find the frequency (f) from the angular frequency (w)

Using the relationship w = 2 * pi * f, we can find the frequency (f) by rearranging the equation: f = w / (2 * pi) Plug in the value of w (2) to get: f = \( \frac{2}{2 * \pi} \) f = \( \frac{1}{\pi} \) Hz
04

Identify the correct option

Now that we have calculated the wavelength (λ) and frequency (f) as \( \frac{\pi}{20} \) meters and \( \frac{1}{\pi} \) Hz, respectively, we can compare these values to the given options: (A) \( (\frac{\pi}{5})m; 0.12 Hz \) (B) \( (\frac{\pi}{10})m; 0.24 Hz \) (C) \( (\frac{\pi}{40})m; 0.48 Hz \) (D) \( (\frac{\pi}{20})m; 0.32 Hz \) The correct option is (D), with a wavelength of \( \frac{\pi}{20} \) meters and a frequency of 0.32 Hz (approximately equal to \( \frac{1}{\pi} \) Hz).

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Most popular questions from this chapter

A rectangular block having mass \(\mathrm{m}\) and cross sectional area A is floating in a liquid having density \(\rho\). If this block in its equilibrium position is given a small vertical displacement, its starts oscillating with periodic time \(\mathrm{T}\). Then in this case \(\ldots \ldots\) (A) \(\mathrm{T} \propto(1 / \sqrt{\mathrm{m}})\) (B) \(T \propto \sqrt{\rho}\) (C) \(\mathrm{T} \propto(1 / \sqrt{\mathrm{A}})\) (D) \(\mathrm{T} \propto(1 / \sqrt{\rho})\)

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