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Figure shows the wave \mathrm{y=A \sin (\omega-k x)}. What is the magnitude of slope of curve at B ?

 

Option: 1

\mathrm{\frac{w}{A}}


Option: 2

\mathrm{\frac{k}{A}}


Option: 3

\mathrm{kA}


Option: 4

\mathrm{wA}


Answers (1)

best_answer

\therefore Particle velocity is maximum at B and is given by -

\mathrm{\frac{d y}{d t}=\left(V_\rho\right)_{\text {max }}=\omega_{\mathrm{A}}}

Also wave velocity is \mathrm{\frac{d x}{d t}=v=\frac{\omega}{k}}

So slope, \mathrm{\begin{aligned} & \frac{d y}{d x}=\frac{\left(V_p\right)_{\text {max }}}{d t}=v=\frac{\omega}{k} \\ & \frac{d y}{d x}=\frac{\left(V_p\right)_{\text {max }}}{d t}=k \cdot A \end{aligned}}

Posted by

shivangi.shekhar

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