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  A transverse wave is represented by

y= \frac{10}{\pi }\sin \left ( \frac{2\pi }{T} t-\frac{2\pi }{\lambda }x\right )

For what value of the wavelength the wave velocity is twice the maximum particle velocity ?

Option: 1

40 cm


Option: 2

20 cm


Option: 3

10 cm


Option: 4

60 cm


Answers (1)

we are learned

Relation between phase velocity and wave speed -

V_{P}= -V\: \frac{dy}{dx}
 

- wherein

V_{P}= particle velocity

V = wave velocity

\frac{dy}{dx}= slope of curve

 

 

y=\frac{10}{\pi }\cdot \sin \left ( \frac{2\pi }{\tau }t-\frac{2\pi }{\lambda }x\right )

particle velocity= \frac{dy}{dt}=\left ( \frac{10}{\pi } \right )\cdot \left ( \frac{2\pi }{\tau } \right )\cdot \cos \left ( \frac{2\pi }{\tau }t-\frac{2\pi }{\lambda }x \right )

=\frac{20}{\tau }\cdot \cos \left ( \frac{2\pi }{\tau }\cdot t-\frac{2\pi }{\lambda \cdot x} \right )

\left ( \frac{dy}{dt} \right )_{max}=\frac{20}{\tau }

wave velocity =\frac{w}{k}=\frac{2\pi }{\tau \cdot \frac{2\pi }{\lambda }}=\frac{\lambda }{\tau }

\therefore\frac{\lambda }{\tau }=2\cdot \frac{20}{\tau }\Rightarrow \Rightarrow \lambda =40cm 

Posted by

Sumit Saini

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