Question

A transverse harmonic wave on a string is described by y=[5cos(2πt + π/4)] m. All parameters are in its SI units. Find its amplitude, velocity and acceleration at 0.125s

Answer 1

Dear Student
$$\begin{gathered} Givenequation \\ y=5\cos \left(2\mathrm{\pi t}+\frac{\mathrm{\pi }}{4}\right) \\ Wecomparewiths\tan dardequation,x\left(t\right)=A\cos (\omega t+\phi ) \\ SoamplitudeA=5 \\ Velocity \\ Velocity,v=\frac{dy}{dt}=\frac{d}{dt}[5\cos \left(2\mathrm{\pi t}+\frac{\mathrm{\pi }}{4}\right)] \\ v=–5.0\times 2\mathrm{\pi sin}\left(2\mathrm{\pi t}+\frac{\mathrm{\pi }}{4}\right) \\ \mathrm{At}\mathrm{t}=1.25\mathrm{s} \\ \mathrm{v}=–5.0\times 2\mathrm{\pi sin}\left(2\mathrm{\pi }\times 1.25+\frac{\mathrm{\pi }}{4}\right) \\ \mathrm{v}=-5\times 2\mathrm{\pi }\left(0.7071\right)=-22.22\mathrm{m}/\mathrm{s} \\ \mathrm{Acceleration} \\ \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}[–10\mathrm{\pi sin}\left(2\mathrm{\pi t}+\frac{\mathrm{\pi }}{4}\right)] \\ \mathrm{a}=-20{\mathrm{\pi }}^2\cos \left(2\mathrm{\pi t}+\frac{\mathrm{\pi }}{4}\right) \\ \mathrm{a}=-4{\mathrm{\pi }}^2\left[5\cos \left(2\mathrm{\pi t}+\frac{\mathrm{\pi }}{4}\right)\right] \\ \mathrm{a}=-4{\mathrm{\pi }}^2\left[5\cos \left(2\mathrm{\pi }\times 1.25+\frac{\mathrm{\pi }}{4}\right)\right] \\ \mathrm{a}=-4{\mathrm{\pi }}^2\left[5\left(-0.7071\right)\right] \\ \mathrm{a}=139.56\mathrm{m}/\mathrm{s} \end{gathered}$$
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