A uniform string suspended vertically. A transverse pulse is created at top most of string. Then

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14.Waves and Sound

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A uniform string suspended vertically. A transverse pulse is created at the top most of the string. Then

A

speed of pulse remains constant

B

the speed of the pulse decreases with constant rate as pulse moves downward.

C

the speed of the pulse decreases with increasing rate as pulse moves downward

D

the speed of the pulse increases with constant rate as pulse moves downward

Solution

Tension at $\mathrm{P}=\frac{\mathrm{m}}{\mathrm{L}} \mathrm{gx}$

${{\rm{v}}_\omega } = \sqrt {{\rm{gx}}} $

${a_\omega } = {v_\omega }\frac{{d{v_\omega }}}{{dx}} = g/2$

$\therefore \quad(2)$

Standard 11

Physics

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