The displacement equation of a travelling sound wave along x axis is given by s=5×10−5sin(500t−2x), where s and x are in meters and t in seconds. Choose the correct option(s) among the following.
A
Ratio of displacement amplitude of the particles to the wavelength of wave is 1.1×10−5.
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B
Ratio of displacement amplitude of the particles to the wavelength of wave is 1.59×10−5.
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C
Ratio of the velocity amplitude of the particles to the wave speed is 2×10−4.
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D
Ratio of the velocity amplitude of the particles to the wave speed is 10−4.
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Solution
The correct options are B Ratio of displacement amplitude of the particles to the wavelength of wave is 1.59×10−5. D Ratio of the velocity amplitude of the particles to the wave speed is 10−4. Given that, s=5×10−5sin(500t−2x) y=Asin(ωt−kx) Comparing the above equations gives s0=5×10−5 m ω=500s−1 k=2m−1 Now, s0λ=s0(2πk)=s0k2π =5×10−5×22π =1.59×10−5
Wave velocity v=ωk=5002m/s Velocity of the particle (vp) is given by dsdt=ddt(5×10−5sin(500t−2x)) ⇒vp=5×10−5×500cos(500t−2x)
∴(vp)maxvwave=5×500×10−55002 =2×5×10−5 =10−4 Thus, options (b) and (d) are correct.