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Q.

A travelling wave in a string has speed 5 cm/s in -ve x direction its amplitude is 10 mm and wavelength 1 m. At a particular time a point P has displacement $5 \sqrt{3} mm$ . Find the velocity vector of point P ?

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a

$- \frac{π}{20} \hat{j} cm / s$

b

$\frac{π}{20} \hat{j} m / s$

c

$+ \frac{2 π}{35} \hat{i} cm / s$

d

$- \frac{π}{20} \hat{i} cm / s$

answer is C.

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Detailed Solution

$\begin{array}{l} f = \frac{v}{λ} = \frac{5}{100} = \frac{1}{20} \Rightarrow ω = 2 πf = \frac{π}{10} \\ \Rightarrow v_{p} = ω \sqrt{A^{2} - y^{2}} = \frac{π}{10} \sqrt{1^{2} - {(5 \sqrt{3} \times 10^{- 1})}^{2}} \\ \Rightarrow v_{p} = \frac{π}{20} cm / s; as v_{p} = - v (slope) \\ \Rightarrow v_{p} = - \frac{π}{20} \hat{j} cm / s \end{array}$

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