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Small amplitude progressive wave in a stretched string has a speed of 100 cms^-1, and frequency 100 Hz. The phase difference between two points 2.75 cm apart on the string, in radians, is

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$11 \frac{π}{2}$

$\frac{3 π}{8}$

$\frac{π}{4}$

answer is B.

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Detailed Solution

$\begin{array}{l} v = 1 m / s, v = 100 Hz \\ λ = \frac{v}{v} = \frac{1}{100} m = 1 cm \\ Δ ϕ = \frac{2 π}{λ} \times 2.75 rad = 5.5 π rad = \frac{11 π}{2} rad \end{array}$

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