A wave is described by the equation y = (0.001 mm) [(50 s^ -1 )t + (2.0 m^ -1 )x] .

Correct answer: C. The amplitude is 0.001 mm The given wave equation is y = (0.001 mm) [(50 s^ -1 )t + (2.0 m^ -1 )x] Comparing this with the standard wave equation y = A (ω t + kx) , we get: Amplitude A = 0.001 mm Angular frequency ω = 50 rad s^ -1 Wave number k = 2.0 m^ -1 The wavelength is given by λ = (2π)/(k) = (2π)/(2.0) = π m The frequency is given by f = (ω)/(2π) = (50)/(2π) = (25)/(π) Hz The wave velocity is given by v = (ω)/(k) = (50)/(2.0) = 25 m s^ -1 Thus, the frequency is 25/π Hz and the amplitude is 0.001 mm .

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Wave Motion and Waves on a String

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Wave Motion and Waves on a String

A wave is described by the equation y = (0.001 mm) [(50 s^ -1 )t +...

A wave is described by the equation

y = (0.001 \text{ mm}) \sin[(50 \text{ s}^{-1})t + (2.0 \text{ m}^{-1})x]

Options

AThe wavelength is $2.0 \text{ m}$
BThe frequency is $25/\pi \text{ Hz}$
CThe amplitude is $0.001 \text{ mm}$
DThe wave velocity is $100 \text{ m s}^{-1}$

Correct answer

C. The amplitude is $0.001 \text{ mm}$

Step-by-step solution

The given wave equation is

y = (0.001 \text{ mm}) \sin[(50 \text{ s}^{-1})t + (2.0 \text{ m}^{-1})x]

Comparing this with the standard wave equation

y = A \sin(\omega t + kx)

, we get:

Amplitude

A = 0.001 \text{ mm}

Angular frequency

\omega = 50 \text{ rad s}^{-1}

Wave number

k = 2.0 \text{ m}^{-1}

The wavelength is given by

\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{2.0} = \pi \text{ m}

The frequency is given by

f = \dfrac{\omega}{2\pi} = \dfrac{50}{2\pi} = \dfrac{25}{\pi} \text{ Hz}

The wave velocity is given by

v = \dfrac{\omega}{k} = \dfrac{50}{2.0} = 25 \text{ m s}^{-1}

Thus, the frequency is

25/\pi \text{ Hz}

and the amplitude is

0.001 \text{ mm}

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